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I was trying out the code for level order traversal in a binary search tree using a queue and I don't know why I'm getting no output when I try to print the level order. Someone please help!
Link to my code.
#include <stdio.h>
#include <stdlib.h>
typedef struct
{
int data;
struct Node *right, *left;
}Node;
typedef struct
{
int front, rear, size;
unsigned capacity;
Node* arr[100];
}Queue;
Queue* createQ()
{
Queue* q = (Queue*) malloc(sizeof(Queue));
q->capacity = 100;
q->front = q->size = 0;
q->rear = q->capacity - 1;
return q;
}
int isEmpty(Queue* q)
{
return (q->size == 0);
}
int isFull(Queue* q)
{
return (q->size == q->capacity);
}
void enqueue(Queue* q, Node* item)
{
if (isFull(q))
return;
q->rear = (q->rear + 1)%q->capacity;
q->arr[q->rear] = item;
q->size = q->size + 1;
}
Node* dequeue(Queue* q)
{
if (isEmpty(q))
return NULL;
Node* item = q->arr[q->front];
q->front = (q->front + 1)%q->capacity;
q->size = q->size - 1;
return item;
}
Node* create(Node* root, int data)
{
if(root==NULL)
{
root = (Node*)malloc(sizeof(Node));
root->data = data;
root->left = root->right = NULL;
return root;
}
else
{
if(data>root->data)
{
root->right = create(root->right,data);
}
else
{
root->left = create(root->left,data);
}
return root;
}
}
void levelorder(Node* root)
{
if(root==NULL) return;
else
{
Queue* q = createQ();
enqueue(q,root);
while(isEmpty(q))
{
Node* temp = dequeue(q);
printf("%d ",temp->data);
if(temp->right)
enqueue(q,temp->right);
else if(temp->left);
enqueue(q,temp->left);
}
}
}
void main()
{
int data;
Node *root = NULL;
for(;;)
{
printf("\nEnter the elements (-1 to stop) : ");
scanf("%d",&data);
if(data==-1) break;
root = create(root,data);
}
printf("\nPrinting the elemtents in level order : ");
levelorder(root);
}
You have several bugs in your levelorder function. Please check the corrections below. I've kept your buggy line in side comment.
void levelorder(Node* root)
{
if(root==NULL) return;
else
{
Queue* q = createQ();
enqueue(q,root);
while(!isEmpty(q)) // not while(isEmpty(q)) you have to check untill Queue is non-empty
{
Node* temp = dequeue(q);
printf("%d ",temp->data);
if(temp->right)
enqueue(q,temp->right);
if(temp->left) // not else if(temp->left); there can be right node and left node as well.. no semicolon
enqueue(q,temp->left);
}
}
}
Rest of the code is okay. You should practice debugging yourself using some well-known IDE.
In line 83, while(isEmpty(q)) should be while(!isEmpty(q)) because we would want to run the loop while the Queue is non-empty.
I hope I got the bug right.
Have a recursive remove function for BST that isn't zeroing a pointer to a leaf node.
bool removeNode(Node* tree, int key)
{
bool removed = false;
if (tree)
{
if (key < tree->key)
{
removeNode(tree->left, key);
}
else if (key > tree->key)
{
removeNode(tree->right, key);
}
else // this node is the key
{
if (!tree->left && !tree->right) // leaf
{
free(tree);
tree = 0;
}
else if (!tree->left)
{
*tree = *tree->right;
}
else if (!tree->right)
{
*tree = *tree->left;
}
else // this node has 2 children
{
Node* paux = tree->left;
if (paux->right)
{
while (paux->right)
{
paux = paux->right;
}
}
tree->key = paux->key;
tree->left = paux->left;
}
removed = true;
}
}
return removed;
}
I use
Node* node = (Node*)malloc(sizeof(Node));
to allocate memory. My leaf's address is correctly zeroed, but when it returns to the previous call, the address remains still the same. If the address was zeroed, why does it return to the previous value? The change should have affected the pointer... shouldn't it?
Data structures and related functions:
typedef struct Node
{
int key;
struct Node* left;
struct Node* right;
} Node;
// init a binary tree
void init(Node** tree, int key)
{
*tree = (Node*) malloc(sizeof(Node));
(*tree)->key = key;
(*tree)->left = 0;
(*tree)->right = 0;
}
// insert at binary tree
bool insert(Node** tree, int key)
{
bool inserted = false;
if (!*tree)
{
init(&*tree, key);
}
else
{
Node* node = (Node*)malloc(sizeof(Node));
node->key = key;
node->left = 0;
node->right = 0;
Node* paux = *tree;
Node* root = paux;
while (paux != 0)
{
root = paux;
if (key < paux->key)
{
paux = paux->left;
}
else
{
paux = paux->right;
}
}
paux = node;
if (key < root->key)
{
root->left = paux;
}
else
{
root->right = paux;
}
inserted = true;
}
return inserted;
}
void print(Node* tree)
{
if (tree != 0)
{
printf("%d ", tree->key);
print(tree->left);
print(tree->right);
}
The problem you encounter happens because the function is not able to modify the input pointer. The pointer itself is passed by value.
Think of a function that increments an integer. The trivial implementation will not work as the argument is passed "by value".
void inc(int x)
{
x++;
}
you can fix this sample by passing it by reference (a pointer)
void inc(int *x)
{
(*x)++;
}
so, if you want to be able to nullify the pointer from within the function pass a pointer to the pointer:
bool removeNode(Node **tree, int key)
As you want to be able to change tree inside the function you need a double pointer.
Instead of
bool removeNode(Node* tree, int key)
you need
bool removeNode(Node** tree, int key)
^^
That also means that you need to change the way you access tree inside the function.
Also see: https://stackoverflow.com/a/39436538/4386427 - it is the same problem
removeNode() has its own copy of tree, so changes to the actual value of tree itself are not visible outside removeNode(). One option is to make tree a Node **, as others have suggested. Another option is to refactor so that you have the parent's pointer available at the point where you want to free the node. For example, the parent could be passed as another parameter to removeNode(), or your code could test one level deeper while processing. Then you could do free(parent->left); parent->left=0; or some such.
I wonder if there exists some logic to reverse a singly-linked list using only two pointers.
The following is used to reverse the single linked list using three pointers namely p, q, r:
struct node {
int data;
struct node *link;
};
void reverse() {
struct node *p = first,
*q = NULL,
*r;
while (p != NULL) {
r = q;
q = p;
p = p->link;
q->link = r;
}
first = q;
}
Is there any other alternate to reverse the linked list? What would be the best logic to reverse a singly linked list, in terms of time complexity?
Any alternative? No, this is as simple as it gets, and there's no fundamentally-different way of doing it. This algorithm is already O(n) time, and you can't get any faster than that, as you must modify every node.
It looks like your code is on the right track, but it's not quite working in the form above. Here's a working version:
#include <stdio.h>
typedef struct Node {
char data;
struct Node* next;
} Node;
void print_list(Node* root) {
while (root) {
printf("%c ", root->data);
root = root->next;
}
printf("\n");
}
Node* reverse(Node* root) {
Node* new_root = 0;
while (root) {
Node* next = root->next;
root->next = new_root;
new_root = root;
root = next;
}
return new_root;
}
int main() {
Node d = { 'd', 0 };
Node c = { 'c', &d };
Node b = { 'b', &c };
Node a = { 'a', &b };
Node* root = &a;
print_list(root);
root = reverse(root);
print_list(root);
return 0;
}
I hate to be the bearer of bad news but I don't think your three-pointer solution actually works. When I used it in the following test harness, the list was reduced to one node, as per the following output:
==========
4
3
2
1
0
==========
4
==========
You won't get better time complexity than your solution since it's O(n) and you have to visit every node to change the pointers, but you can do a solution with only two extra pointers quite easily, as shown in the following code:
#include <stdio.h>
// The list element type and head.
struct node {
int data;
struct node *link;
};
static struct node *first = NULL;
// A reverse function which uses only two extra pointers.
void reverse() {
// curNode traverses the list, first is reset to empty list.
struct node *curNode = first;
first = NULL;
// Until no more in list, insert current before first and advance.
while (curNode != NULL) {
// Need to save next node since we're changing the current.
struct node *nxtNode = curNode->link;
// Insert at start of new list.
curNode->link = first;
first = curNode;
// Advance to next.
curNode = nxtNode;
}
}
// Code to dump the current list.
static void dumpNodes() {
struct node *curNode = first;
printf ("==========\n");
while (curNode != NULL) {
printf ("%d\n", curNode->data);
curNode = curNode->link;
}
}
// Test harness main program.
int main (void) {
int i;
struct node *newnode;
// Create list (using actually the same insert-before-first
// that is used in reverse function.
for (i = 0; i < 5; i++) {
newnode = malloc (sizeof (struct node));
newnode->data = i;
newnode->link = first;
first = newnode;
}
// Dump list, reverse it, then dump again.
dumpNodes();
reverse();
dumpNodes();
printf ("==========\n");
return 0;
}
This code outputs:
==========
4
3
2
1
0
==========
0
1
2
3
4
==========
which I think is what you were after. It can actually do this since, once you've loaded up first into the pointer traversing the list, you can re-use first at will.
#include <stddef.h>
typedef struct Node {
struct Node *next;
int data;
} Node;
Node * reverse(Node *cur) {
Node *prev = NULL;
while (cur) {
Node *temp = cur;
cur = cur->next; // advance cur
temp->next = prev;
prev = temp; // advance prev
}
return prev;
}
Here's the code to reverse a singly linked list in C.
And here it is pasted below:
// reverse.c
#include <stdio.h>
#include <assert.h>
typedef struct node Node;
struct node {
int data;
Node *next;
};
void spec_reverse();
Node *reverse(Node *head);
int main()
{
spec_reverse();
return 0;
}
void print(Node *head) {
while (head) {
printf("[%d]->", head->data);
head = head->next;
}
printf("NULL\n");
}
void spec_reverse() {
// Create a linked list.
// [0]->[1]->[2]->NULL
Node node2 = {2, NULL};
Node node1 = {1, &node2};
Node node0 = {0, &node1};
Node *head = &node0;
print(head);
head = reverse(head);
print(head);
assert(head == &node2);
assert(head->next == &node1);
assert(head->next->next == &node0);
printf("Passed!");
}
// Step 1:
//
// prev head next
// | | |
// v v v
// NULL [0]->[1]->[2]->NULL
//
// Step 2:
//
// prev head next
// | | |
// v v v
// NULL<-[0] [1]->[2]->NULL
//
Node *reverse(Node *head)
{
Node *prev = NULL;
Node *next;
while (head) {
next = head->next;
head->next = prev;
prev = head;
head = next;
}
return prev;
}
Robert Sedgewick, "Algorithms in C", Addison-Wesley, 3rd Edition, 1997, [Section 3.4]
In case that is not a cyclic list ,hence NULL is the last link.
typedef struct node* link;
struct node{
int item;
link next;
};
/* you send the existing list to reverse() and returns the reversed one */
link reverse(link x){
link t, y = x, r = NULL;
while(y != NULL){
t = y->next;
y-> next = r;
r = y;
y = t;
}
return r;
}
Yes. I'm sure you can do this the same way you can swap two numbers without using a third. Simply cast the pointers to a int/long and perform the XOR operation a couple of times. This is one of those C tricks that makes for a fun question, but doesn't have any practical value.
Can you reduce the O(n) complexity? No, not really. Just use a doubly linked list if you think you are going to need the reverse order.
Just for fun (although tail recursion optimization should stop it eating all the stack):
Node* reverse (Node *root, Node *end) {
Node *next = root->next;
root->next = end;
return (next ? reverse(next, root) : root);
}
root = reverse(root, NULL);
You need a track pointer which will track the list.
You need two pointers :
first pointer to pick first node.
second pointer to pick second node.
Processing :
Move Track Pointer
Point second node to first node
Move First pointer one step, by assigning second pointer to one
Move Second pointer one step, By assigning Track pointer to second
Node* reverselist( )
{
Node *first = NULL; // To keep first node
Node *second = head; // To keep second node
Node *track = head; // Track the list
while(track!=NULL)
{
track = track->next; // track point to next node;
second->next = first; // second node point to first
first = second; // move first node to next
second = track; // move second node to next
}
track = first;
return track;
}
How about the more readable:
Node *pop (Node **root)
{
Node *popped = *root;
if (*root) {
*root = (*root)->next;
}
return (popped);
}
void push (Node **root, Node *new_node)
{
new_node->next = *root;
*root = new_node;
}
Node *reverse (Node *root)
{
Node *new_root = NULL;
Node *next;
while ((next = pop(&root))) {
push (&new_root, next);
}
return (new_root);
}
To swap two variables without the use of a temporary variable,
a = a xor b
b = a xor b
a = a xor b
fastest way is to write it in one line
a = a ^ b ^ (b=a)
Similarly,
using two swaps
swap(a,b)
swap(b,c)
solution using xor
a = a^b^c
b = a^b^c
c = a^b^c
a = a^b^c
solution in one line
c = a ^ b ^ c ^ (a=b) ^ (b=c)
b = a ^ b ^ c ^ (c=a) ^ (a=b)
a = a ^ b ^ c ^ (b=c) ^ (c=a)
The same logic is used to reverse a linked list.
typedef struct List
{
int info;
struct List *next;
}List;
List* reverseList(List *head)
{
p=head;
q=p->next;
p->next=NULL;
while(q)
{
q = (List*) ((int)p ^ (int)q ^ (int)q->next ^ (int)(q->next=p) ^ (int)(p=q));
}
head = p;
return head;
}
Here's a simpler version in Java. It does use only two pointers curr & prev
public void reverse(Node head) {
Node curr = head, prev = null;
while (head.next != null) {
head = head.next; // move the head to next node
curr.next = prev; //break the link to the next node and assign it to previous
prev = curr; // we are done with previous, move it to next node
curr = head; // current moves along with head
}
head.next = prev; //for last node
}
Work out the time complexity of the algorithm you are using now and it should be obvious that it can not be improved.
I don't understand why there is need to return head as we are passing it as argument. We are passing head of the link list then we can update also. Below is simple solution.
#include<stdio.h>
#include<conio.h>
struct NODE
{
struct NODE *next;
int value;
};
typedef struct NODE node;
void reverse(node **head);
void add_end(node **head,int val);
void alloc(node **p);
void print_all(node *head);
void main()
{
node *head;
clrscr();
head = NULL;
add_end( &head, 1 );
add_end( &head, 2 );
add_end( &head, 3 );
print_all( head );
reverse( &head );
print_all( head );
getch();
}
void alloc(node **p)
{
node *temp;
temp = (node *) malloc( sizeof(node *) );
temp->next = NULL;
*p = temp;
}
void add_end(node **head,int val)
{
node *temp,*new_node;
alloc(&new_node);
new_node->value = val;
if( *head == NULL )
{
*head = new_node;
return;
}
for(temp = *head;temp->next!=NULL;temp=temp->next);
temp->next = new_node;
}
void print_all(node *head)
{
node *temp;
int index=0;
printf ("\n\n");
if (head == NULL)
{
printf (" List is Empty \n");
return;
}
for (temp=head; temp != NULL; temp=temp->next,index++)
printf (" %d ==> %d \n",index,temp->value);
}
void reverse(node **head)
{
node *next,*new_head;
new_head=NULL;
while(*head != NULL)
{
next = (*head)->next;
(*head)->next = new_head;
new_head = (*head);
(*head) = next;
}
(*head)=new_head;
}
#include <stdio.h>
#include <malloc.h>
tydef struct node
{
int info;
struct node *link;
} *start;
void main()
{
rev();
}
void rev()
{
struct node *p = start, *q = NULL, *r;
while (p != NULL)
{
r = q;
q = p;
p = p->link;
q->link = r;
}
start = q;
}
curr = head;
prev = NULL;
while (curr != NULL) {
next = curr->next; // store current's next, since it will be overwritten
curr->next = prev;
prev = curr;
curr = next;
}
head = prev; // update head
No, nothing faster than the current O(n) can be done. You need to alter every node, so time will be proportional to the number of elements anyway and that's O(n) you already have.
Using two pointers while maintaining time complexity of O(n), the fastest achievable, might only be possible through number casting of pointers and swapping their values. Here is an implementation:
#include <stdio.h>
typedef struct node
{
int num;
struct node* next;
}node;
void reverse(node* head)
{
node* ptr;
if(!head || !head->next || !head->next->next) return;
ptr = head->next->next;
head->next->next = NULL;
while(ptr)
{
/* Swap head->next and ptr. */
head->next = (unsigned)(ptr =\
(unsigned)ptr ^ (unsigned)(head->next =\
(unsigned)head->next ^ (unsigned)ptr)) ^ (unsigned)head->next;
/* Swap head->next->next and ptr. */
head->next->next = (unsigned)(ptr =\
(unsigned)ptr ^ (unsigned)(head->next->next =\
(unsigned)head->next->next ^ (unsigned)ptr)) ^ (unsigned)head->next->next;
}
}
void add_end(node* ptr, int n)
{
while(ptr->next) ptr = ptr->next;
ptr->next = malloc(sizeof(node));
ptr->next->num = n;
ptr->next->next = NULL;
}
void print(node* ptr)
{
while(ptr = ptr->next) printf("%d ", ptr->num);
putchar('\n');
}
void erase(node* ptr)
{
node *end;
while(ptr->next)
{
if(ptr->next->next) ptr = ptr->next;
else
{
end = ptr->next;
ptr->next = NULL;
free(end);
}
}
}
void main()
{
int i, n = 5;
node* dummy_head;
dummy_head->next = NULL;
for(i = 1; i <= n ; ++i) add_end(dummy_head, i);
print(dummy_head);
reverse(dummy_head);
print(dummy_head);
erase(dummy_head);
}
I have a slightly different approach. I wanted to make use of the existing functions (like insert_at(index), delete_from(index)) to reverse the list (something like a right shift operation). The complexity is still O(n) but the advantage is more reused code. Have a look at another_reverse() method and let me know what you all think.
#include <stdio.h>
#include <stdlib.h>
struct node {
int data;
struct node* next;
};
struct node* head = NULL;
void printList(char* msg) {
struct node* current = head;
printf("\n%s\n", msg);
while (current != NULL) {
printf("%d ", current->data);
current = current->next;
}
}
void insert_beginning(int data) {
struct node* newNode = (struct node*) malloc(sizeof(struct node));
newNode->data = data;
newNode->next = NULL;
if (head == NULL)
{
head = newNode;
} else {
newNode->next = head;
head = newNode;
}
}
void insert_at(int data, int location) {
struct node* newNode = (struct node*) malloc(sizeof(struct node));
newNode->data = data;
newNode->next = NULL;
if (head == NULL)
{
head = newNode;
}
else {
struct node* currentNode = head;
int index = 0;
while (currentNode != NULL && index < (location - 1)) {
currentNode = currentNode->next;
index++;
}
if (currentNode != NULL)
{
if (location == 0) {
newNode->next = currentNode;
head = newNode;
} else {
newNode->next = currentNode->next;
currentNode->next = newNode;
}
}
}
}
int delete_from(int location) {
int retValue = -1;
if (location < 0 || head == NULL)
{
printf("\nList is empty or invalid index");
return -1;
} else {
struct node* currentNode = head;
int index = 0;
while (currentNode != NULL && index < (location - 1)) {
currentNode = currentNode->next;
index++;
}
if (currentNode != NULL)
{
// we've reached the node just one prior to the one we want to delete
if (location == 0) {
if (currentNode->next == NULL)
{
// this is the only node in the list
retValue = currentNode->data;
free(currentNode);
head = NULL;
} else {
// the next node should take its place
struct node* nextNode = currentNode->next;
head = nextNode;
retValue = currentNode->data;
free(currentNode);
}
} // if (location == 0)
else {
// the next node should take its place
struct node* nextNode = currentNode->next;
currentNode->next = nextNode->next;
if (nextNode != NULL
) {
retValue = nextNode->data;
free(nextNode);
}
}
} else {
printf("\nInvalid index");
return -1;
}
}
return retValue;
}
void another_reverse() {
if (head == NULL)
{
printf("\nList is empty\n");
return;
} else {
// get the tail pointer
struct node* tailNode = head;
int index = 0, counter = 0;
while (tailNode->next != NULL) {
tailNode = tailNode->next;
index++;
}
// now tailNode points to the last node
while (counter != index) {
int data = delete_from(index);
insert_at(data, counter);
counter++;
}
}
}
int main(int argc, char** argv) {
insert_beginning(4);
insert_beginning(3);
insert_beginning(2);
insert_beginning(1);
insert_beginning(0);
/* insert_at(5, 0);
insert_at(4, 1);
insert_at(3, 2);
insert_at(1, 1);*/
printList("Original List\0");
//reverse_list();
another_reverse();
printList("Reversed List\0");
/* delete_from(2);
delete_from(2);*/
//printList();
return 0;
}
using 2-pointers....bit large but simple and efficient
void reverse()
{
int n=0;
node *temp,*temp1;
temp=strptr;
while(temp->next!=NULL)
{
n++; //counting no. of nodes
temp=temp->next;
}
// we will exchange ist by last.....2nd by 2nd last so.on....
int i=n/2;
temp=strptr;
for(int j=1;j<=(n-i+1);j++)
temp=temp->next;
// i started exchanging from in between ....so we do no have to traverse list so far //again and again for exchanging
while(i>0)
{
temp1=strptr;
for(int j=1;j<=i;j++)//this loop for traversing nodes before n/2
temp1=temp1->next;
int t;
t=temp1->info;
temp1->info=temp->info;
temp->info=t;
i--;
temp=temp->next;
//at the end after exchanging say 2 and 4 in a 5 node list....temp will be at 5 and we will traverse temp1 to ist node and exchange ....
}
}
#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
struct node
{
int data;
struct node *link;
};
struct node *first=NULL,*last=NULL,*next,*pre,*cur,*temp;
void create()
{
cur=(struct node*) malloc(sizeof(struct node));
printf("enter first data to insert");
scanf("%d",&cur->data);
first=last=cur;
first->link=NULL;
}
void insert()
{
int pos,c;
cur=(struct node*) malloc(sizeof(struct node));
printf("enter data to insert and also its position");
scanf("%d%d",&cur->data,&pos);
if(pos==1)
{
cur->link=first;
first=cur;
}
else
{
c=1;
next=first;
while(c<pos)
{
pre=next;
next=next->link;
c++;
}
if(pre==NULL)
{
printf("Invalid position");
}
else
{
cur->link=pre->link;
pre->link=cur;
}
}
}
void display()
{
cur=first;
while(cur!=NULL)
{
printf("data= %d\t address= %u\n",cur->data,cur);
cur=cur->link;
}
printf("\n");
}
void rev()
{
pre=NULL;
cur=first;
while(cur!=NULL)
{
next=cur->link;
cur->link=pre;
pre=cur;
cur=next;
}
first=pre;
}
void main()
{
int choice;
clrscr();
do
{
printf("Options are: -\n1:Create\n2:Insert\n3:Display\n4:Reverse\n0:Exit\n");
printf("Enter your choice: - ");
scanf("%d",&choice);
switch(choice)
{
case 1:
create();
break;
case 2:
insert();
break;
case 3:
display();
break;
case 4:
rev();
break;
case 0:
exit(0);
default:
printf("wrong choice");
}
}
while(1);
}
Yes there is a way using only two pointers. That is by creating new linked list where the first node is the first node of the given list and second node of the first list is added at the start of the new list and so on.
Here is my version:
void reverse(ListElem *&head)
{
ListElem* temp;
ListElem* elem = head->next();
ListElem* prev = head;
head->next(0);
while(temp = elem->next())
{
elem->next(prev);
prev = elem;
elem = temp;
}
elem->next(prev);
head = elem;
}
where
class ListElem{
public:
ListElem(int val): _val(val){}
ListElem *next() const { return _next; }
void next(ListElem *elem) { _next = elem; }
void val(int val){ _val = val; }
int val() const { return _val;}
private:
ListElem *_next;
int _val;
};
I am using java to implement this and approach is test driven development hence test cases are also attached.
The Node class that represent single node -
package com.adnan.linkedlist;
/**
* User : Adnan
* Email : sendtoadnan#gmail.com
* Date : 9/21/13
* Time : 12:02 PM
*/
public class Node {
public Node(int value, Node node){
this.value = value;
this.node = node;
}
private int value;
private Node node;
public int getValue() {
return value;
}
public Node getNode() {
return node;
}
public void setNode(Node node){
this.node = node;
}
}
Service class that takes start node as input and reserve it without using extra space.
package com.adnan.linkedlist;
/**
* User : Adnan
* Email : sendtoadnan#gmail.com
* Date : 9/21/13
* Time : 11:54 AM
*/
public class SinglyLinkedListReversal {
private static final SinglyLinkedListReversal service
= new SinglyLinkedListReversal();
public static SinglyLinkedListReversal getService(){
return service;
}
public Node reverse(Node start){
if (hasOnlyNodeInLinkedList(start)){
return start;
}
Node firstNode, secondNode, thirdNode;
firstNode = start;
secondNode = firstNode.getNode();
while (secondNode != null ){
thirdNode = secondNode.getNode();
secondNode.setNode(firstNode);
firstNode = secondNode;
secondNode = thirdNode;
}
start.setNode(null);
return firstNode;
}
private boolean hasOnlyNodeInLinkedList(Node start) {
return start.getNode() == null;
}
}
And The test case that covers above scenario. Please note that you require junit jars. I am using testng.jar; you can use any whatever pleases you..
package com.adnan.linkedlist;
import org.testng.annotations.Test;
import static org.testng.AssertJUnit.assertTrue;
/**
* User : Adnan
* Email : sendtoadnan#gmail.com
* Date : 9/21/13
* Time : 12:11 PM
*/
public class SinglyLinkedListReversalTest {
private SinglyLinkedListReversal reversalService =
SinglyLinkedListReversal.getService();
#Test
public void test_reverseSingleElement() throws Exception {
Node node = new Node(1, null);
reversalService.reverse(node);
assertTrue(node.getNode() == null);
assertTrue(node.getValue() == 1);
}
//original - Node1(1) -> Node2(2) -> Node3(3)
//reverse - Node3(3) -> Node2(2) -> Node1(1)
#Test
public void test_reverseThreeElement() throws Exception {
Node node3 = new Node(3, null);
Node node2 = new Node(2, node3);
Node start = new Node(1, node2);
start = reversalService.reverse(start);
Node test = start;
for (int i = 3; i >=1 ; i -- ){
assertTrue(test.getValue() == i);
test = test.getNode();
}
}
#Test
public void test_reverseFourElement() throws Exception {
Node node4 = new Node(4, null);
Node node3 = new Node(3, node4);
Node node2 = new Node(2, node3);
Node start = new Node(1, node2);
start = reversalService.reverse(start);
Node test = start;
for (int i = 4; i >=1 ; i -- ){
assertTrue(test.getValue() == i);
test = test.getNode();
}
}
#Test
public void test_reverse10Element() throws Exception {
Node node10 = new Node(10, null);
Node node9 = new Node(9, node10);
Node node8 = new Node(8, node9);
Node node7 = new Node(7, node8);
Node node6 = new Node(6, node7);
Node node5 = new Node(5, node6);
Node node4 = new Node(4, node5);
Node node3 = new Node(3, node4);
Node node2 = new Node(2, node3);
Node start = new Node(1, node2);
start = reversalService.reverse(start);
Node test = start;
for (int i = 10; i >=1 ; i -- ){
assertTrue(test.getValue() == i);
test = test.getNode();
}
}
#Test
public void test_reverseTwoElement() throws Exception {
Node node2 = new Node(2, null);
Node start = new Node(1, node2);
start = reversalService.reverse(start);
Node test = start;
for (int i = 2; i >=1 ; i -- ){
assertTrue(test.getValue() == i);
test = test.getNode();
}
}
}
A simple algorithm if you use the linked list as a stack structure:
#include <stdio.h>
#include <stdlib.h>
typedef struct list {
int key;
char value;
struct list* next;
} list;
void print(list*);
void add(list**, int, char);
void reverse(list**);
void deleteList(list*);
int main(void) {
list* head = NULL;
int i=0;
while ( i++ < 26 ) add(&head, i, i+'a');
printf("Before reverse: \n");
print(head);
printf("After reverse: \n");
reverse(&head);
print(head);
deleteList(head);
}
void deleteList(list* l) {
list* t = l;
while ( t != NULL ) {
list* tmp = t;
t = t->next;
free(tmp);
}
}
void print(list* l) {
list* t = l;
while ( t != NULL) {
printf("%d:%c\n", t->key, t->value);
t = t->next;
}
}
void reverse(list** head) {
list* tmp = *head;
list* reversed = NULL;
while ( tmp != NULL ) {
add(&reversed, tmp->key, tmp->value);
tmp = tmp->next;
}
deleteList(*head);
*head = reversed;
}
void add(list** head, int k, char v) {
list* t = calloc(1, sizeof(list));
t->key = k; t->value = v;
t->next = *head;
*head = t;
}
The performance may be affected since additional function call to the add and malloc so the algorithms of address swaps are better but that one actually creates new list so you can use additional options like sort or remove items if you add a callback function as parameter to the reverse.
Here is a slightly different, but simple approach in C++11:
#include <iostream>
struct Node{
Node(): next(NULL){}
Node *next;
std::string data;
};
void printlist(Node* l){
while(l){
std::cout<<l->data<<std::endl;
l = l->next;
}
std::cout<<"----"<<std::endl;
}
void reverse(Node*& l)
{
Node* prev = NULL;
while(l){
auto next = l->next;
l->next = prev;
prev=l;
l=next;
}
l = prev;
}
int main() {
Node s,t,u,v;
s.data = "1";
t.data = "2";
u.data = "3";
v.data = "4";
s.next = &t;
t.next = &u;
u.next = &v;
Node* ptr = &s;
printlist(ptr);
reverse(ptr);
printlist(ptr);
return 0;
}
Output here
Following is one implementation using 2 pointers (head and r)
ListNode * reverse(ListNode* head) {
ListNode *r = NULL;
if(head) {
r = head->next;
head->next = NULL;
}
while(r) {
head = reinterpret_cast<ListNode*>(size_t(head) ^ size_t(r->next));
r->next = reinterpret_cast<ListNode*>(size_t(r->next) ^ size_t(head));
head = reinterpret_cast<ListNode*>(size_t(head) ^ size_t(r->next));
head = reinterpret_cast<ListNode*>(size_t(head) ^ size_t(r));
r = reinterpret_cast<ListNode*>(size_t(r) ^ size_t(head));
head = reinterpret_cast<ListNode*>(size_t(head) ^ size_t(r));
}
return head;
}
here is a little simple solution...
void reverse()
{
node * pointer1 = head->next;
if(pointer1 != NULL)
{
node *pointer2 = pointer1->next;
pointer1->next = head;
head->next = NULL;
head = pointer1;
if(pointer2 != NULL)
{
while(pointer2 != NULL)
{
pointer1 = pointer2;
pointer2 = pointer2->next;
pointer1->next = head;
head = pointer1;
}
pointer1->next = head;
head = pointer1;
}
}
}
You can have solution of this problem with help of only one extra pointer, that has to be static for the reverse function. It's in O(n) complexity.
#include<stdio.h>
#include<stdlib.h>
typedef struct List* List;
struct List {
int val;
List next;
};
List reverse(List list) { /* with recursion and one static variable*/
static List tail;
if(!list || !list->next) {
tail = list;
return tail;
} else {
reverse1(list->next);
list->next->next = list;
list->next = NULL;
return tail;
}
}
As an alternative, you can use recursion-
struct node* reverseList(struct node *head)
{
if(head == NULL) return NULL;
if(head->next == NULL) return head;
struct node* second = head->next;
head->next = NULL;
struct node* remaining = reverseList(second);
second->next = head;
return remaining;
}
class Node {
Node next;
int data;
Node(int item) {
data = item;
next = null;
}
}
public class LinkedList {
static Node head;
//Print LinkedList
public static void printList(Node node){
while(node!=null){
System.out.print(node.data+" ");
node = node.next;
}
System.out.println();
}
//Reverse the LinkedList Utility
public static Node reverse(Node node){
Node new_node = null;
while(node!=null){
Node next = node.next;
node.next = new_node;
new_node = node;
node = next;
}
return new_node;
}
public static void main(String[] args) {
//Creating LinkedList
LinkedList.head = new Node(1);
LinkedList.head.next = new Node(2);
LinkedList.head.next.next = new Node(3);
LinkedList.head.next.next.next = new Node(4);
LinkedList.printList(LinkedList.head);
Node node = LinkedList.reverse(LinkedList.head);
LinkedList.printList(node);
}
}
For practice, I've been working on a compressor which does the find-repeated-parts, make-dictionary, compress-with-huffman codes thing.
It's not really working.
One of the problems is, for some reason my sorting algorithm drops keywords from the dictionary. I think the problem is in the swap routine, but I'm not sure.
( this routine swaps adjacent keywords, with next being current->next ).
I do have a static keyword * head;
void swap(keyword * current, keyword * next) {
keyword * prev = current->prev;
if (prev){
prev->next = next;
next->prev = prev;
} else { /* no prev - current is head */
head = next;
next->prev = 0;
}
current->prev = next;
current->next = next->next;
next->next = current;
}
Spot anything wrong with this?
You don't set next->next->prev.
Getting data-structure implementations correct is notoriously difficult. The way to spot this sort of thing is to work out how many pointers should need updating (6). You're only updating 5, so one must be missing!
I have seen many question related to Linked Lists, so I decide to share my Linked List version, so it can be adapted and/or improved. This version is written in C++, some methods were omitted for saving space:
class Node {
friend class LinkedList;
public:
Node();
Node(const node& krNode);
explicit Node(const int kiData);
virtual ~Node();
Node& operator=(const Node& krNode);
int GetData(void) const {return _idata; }
private: int _iData; Node* _pNetxNode;
}
Node::Node(): _iData(0), _pNextNode(0) {}
Node::Node(const Node& krnode): _iData(krNode._iData), _pNextNode(0) {}
Node::~Node() {}
Node& Node::operator=(const Node& krNode) {_iData = krNode._iData; return *this; }
class LinkedList {
public:
LinkedList();
virtual ~LinkedList();
int GethLenght(void) {return _iSize;}
void Append(Node* pNode);
void Insert(const int kiIndex, node* pNode);
Node* GetItem(const int kiIndex);
int IndexOf(Node* pNode);
void Remove(const int kiInde);
void Swap(constconst int kiIndexA, const int kiIndexB);
void Clear(void);
void Print(void);
protected:
LinkedList(const LinkedList& krLinkedList);
LinkedList& operator=(const LinkedList& krLinkedList);
private:
Node* Detach(const int kiIndex);
Node* _pFirstNode;
Node* _pLastNode;
int _iSize;
}
LinkedList::LinkedList() : _pFirstNode(0), _pLastNode(0), _iSize(0) {}
LinkedList::~LinkedList() { Clear(); }
void LinkedList::Append(Node* pnode) { if(0==_iSize{_pFistrNode = pNode; } else { _pLastNode->_pNextNode = pnode; } _pLastNode = pNode; _iSize++; }
void LinkedList::Insert(const int kiIndex, node* pNode) {
Node* pNext = 0; Node* pPrevious = 0;
if(0==_iSize) { Append(pNode); }
else {
if(0==kiIndex){ pNode->_pNextNode = _pFirstNode; _pFirstNode = pNode; }
else { pPrevious = Getitem(kiIndex-1); pNext=pPrevoius->_pNextnode; pNode->_pNextNode=pNext; pPrevious->_pNextnode=pNode;}
} _iSize++;
}
Node* LinkedList::GetItem(const int kiIndex){
Node* pNode = _pFirstNode; int iCounter = 0;
while(icounter++ != kiIndex) { pNode = pNode->_pNextnode; }
return pNode;
}
int LinkedList::IndexOf(Node* pSearchNode){
node* pNode = _pFirstnode; int iIndex=0;
while(o != pNode) { if(pnode==pSearchNode) { break;} iIdex++; pNode=pnode->_pnextnode; }
if(iIndex ==_iSize) {iIndex = -1;}
return iIndex;
}
void LinkedList::Remove(const int kiIndex){ delete Detach(kiIndex); }
void LinkedList::Swap(const int kiIndexA, const int kiIndexB){
int iLowIndex = 0; int iHighIndex = 0;
if(kiIndex>==kiIndexB) {return;}
iLowIndex = (kiIndexA < kiIndexB) ? kiIndexA : kiIndexB;
iHighIndex = (kiIndexA > kiIndexB) ? kiIndexA : kiIndexB;
Insert(iHighIndex, Detach(iLowIndex));
Insert(iLowIndex, Detach(iHighIndex-1));
}
void LinkedList::Clear(void) {
Node* pNode=_pFirstNode; while(0!=pnode){delete pNode; pNode=pNode-_pNextNode;}
_pFirstnode=0; _pLastnode=0;
}
void LinkedList::Print(void) {
Node* pNode = _pFirstNode;
while(0 != pnode) {printf("%d ", pNode_iData); pNode = pNode->_pNextNode;}
}
Node* LinkedList::Detach(const int kiindex){
Node* pnode = _pFirstnode; Node* pToDetachNode = 0;
if(kiIndex==0){
pToDetachNode = _pFirstNode; _pFirstNode=pnode->_pNextNode; if(1==_iSize){_pLastNode=0;}
}
else{
Node* pPreviousNode = GetItem(kiIndex-1);
pToDetachNode = pPreviousNode->_pNextNode;
if(kiIndex<_iSize){pPreviousNode->_pNextNode=pPreviousNode->_pNextnode; }
else {pPreviousNode->_pNextnode=0; _pLastNode=pPrevoiusNode;}
}_iSize--;
}
I wonder if there exists some logic to reverse a singly-linked list using only two pointers.
The following is used to reverse the single linked list using three pointers namely p, q, r:
struct node {
int data;
struct node *link;
};
void reverse() {
struct node *p = first,
*q = NULL,
*r;
while (p != NULL) {
r = q;
q = p;
p = p->link;
q->link = r;
}
first = q;
}
Is there any other alternate to reverse the linked list? What would be the best logic to reverse a singly linked list, in terms of time complexity?
Any alternative? No, this is as simple as it gets, and there's no fundamentally-different way of doing it. This algorithm is already O(n) time, and you can't get any faster than that, as you must modify every node.
It looks like your code is on the right track, but it's not quite working in the form above. Here's a working version:
#include <stdio.h>
typedef struct Node {
char data;
struct Node* next;
} Node;
void print_list(Node* root) {
while (root) {
printf("%c ", root->data);
root = root->next;
}
printf("\n");
}
Node* reverse(Node* root) {
Node* new_root = 0;
while (root) {
Node* next = root->next;
root->next = new_root;
new_root = root;
root = next;
}
return new_root;
}
int main() {
Node d = { 'd', 0 };
Node c = { 'c', &d };
Node b = { 'b', &c };
Node a = { 'a', &b };
Node* root = &a;
print_list(root);
root = reverse(root);
print_list(root);
return 0;
}
I hate to be the bearer of bad news but I don't think your three-pointer solution actually works. When I used it in the following test harness, the list was reduced to one node, as per the following output:
==========
4
3
2
1
0
==========
4
==========
You won't get better time complexity than your solution since it's O(n) and you have to visit every node to change the pointers, but you can do a solution with only two extra pointers quite easily, as shown in the following code:
#include <stdio.h>
// The list element type and head.
struct node {
int data;
struct node *link;
};
static struct node *first = NULL;
// A reverse function which uses only two extra pointers.
void reverse() {
// curNode traverses the list, first is reset to empty list.
struct node *curNode = first;
first = NULL;
// Until no more in list, insert current before first and advance.
while (curNode != NULL) {
// Need to save next node since we're changing the current.
struct node *nxtNode = curNode->link;
// Insert at start of new list.
curNode->link = first;
first = curNode;
// Advance to next.
curNode = nxtNode;
}
}
// Code to dump the current list.
static void dumpNodes() {
struct node *curNode = first;
printf ("==========\n");
while (curNode != NULL) {
printf ("%d\n", curNode->data);
curNode = curNode->link;
}
}
// Test harness main program.
int main (void) {
int i;
struct node *newnode;
// Create list (using actually the same insert-before-first
// that is used in reverse function.
for (i = 0; i < 5; i++) {
newnode = malloc (sizeof (struct node));
newnode->data = i;
newnode->link = first;
first = newnode;
}
// Dump list, reverse it, then dump again.
dumpNodes();
reverse();
dumpNodes();
printf ("==========\n");
return 0;
}
This code outputs:
==========
4
3
2
1
0
==========
0
1
2
3
4
==========
which I think is what you were after. It can actually do this since, once you've loaded up first into the pointer traversing the list, you can re-use first at will.
#include <stddef.h>
typedef struct Node {
struct Node *next;
int data;
} Node;
Node * reverse(Node *cur) {
Node *prev = NULL;
while (cur) {
Node *temp = cur;
cur = cur->next; // advance cur
temp->next = prev;
prev = temp; // advance prev
}
return prev;
}
Here's the code to reverse a singly linked list in C.
And here it is pasted below:
// reverse.c
#include <stdio.h>
#include <assert.h>
typedef struct node Node;
struct node {
int data;
Node *next;
};
void spec_reverse();
Node *reverse(Node *head);
int main()
{
spec_reverse();
return 0;
}
void print(Node *head) {
while (head) {
printf("[%d]->", head->data);
head = head->next;
}
printf("NULL\n");
}
void spec_reverse() {
// Create a linked list.
// [0]->[1]->[2]->NULL
Node node2 = {2, NULL};
Node node1 = {1, &node2};
Node node0 = {0, &node1};
Node *head = &node0;
print(head);
head = reverse(head);
print(head);
assert(head == &node2);
assert(head->next == &node1);
assert(head->next->next == &node0);
printf("Passed!");
}
// Step 1:
//
// prev head next
// | | |
// v v v
// NULL [0]->[1]->[2]->NULL
//
// Step 2:
//
// prev head next
// | | |
// v v v
// NULL<-[0] [1]->[2]->NULL
//
Node *reverse(Node *head)
{
Node *prev = NULL;
Node *next;
while (head) {
next = head->next;
head->next = prev;
prev = head;
head = next;
}
return prev;
}
Robert Sedgewick, "Algorithms in C", Addison-Wesley, 3rd Edition, 1997, [Section 3.4]
In case that is not a cyclic list ,hence NULL is the last link.
typedef struct node* link;
struct node{
int item;
link next;
};
/* you send the existing list to reverse() and returns the reversed one */
link reverse(link x){
link t, y = x, r = NULL;
while(y != NULL){
t = y->next;
y-> next = r;
r = y;
y = t;
}
return r;
}
Yes. I'm sure you can do this the same way you can swap two numbers without using a third. Simply cast the pointers to a int/long and perform the XOR operation a couple of times. This is one of those C tricks that makes for a fun question, but doesn't have any practical value.
Can you reduce the O(n) complexity? No, not really. Just use a doubly linked list if you think you are going to need the reverse order.
Just for fun (although tail recursion optimization should stop it eating all the stack):
Node* reverse (Node *root, Node *end) {
Node *next = root->next;
root->next = end;
return (next ? reverse(next, root) : root);
}
root = reverse(root, NULL);
You need a track pointer which will track the list.
You need two pointers :
first pointer to pick first node.
second pointer to pick second node.
Processing :
Move Track Pointer
Point second node to first node
Move First pointer one step, by assigning second pointer to one
Move Second pointer one step, By assigning Track pointer to second
Node* reverselist( )
{
Node *first = NULL; // To keep first node
Node *second = head; // To keep second node
Node *track = head; // Track the list
while(track!=NULL)
{
track = track->next; // track point to next node;
second->next = first; // second node point to first
first = second; // move first node to next
second = track; // move second node to next
}
track = first;
return track;
}
How about the more readable:
Node *pop (Node **root)
{
Node *popped = *root;
if (*root) {
*root = (*root)->next;
}
return (popped);
}
void push (Node **root, Node *new_node)
{
new_node->next = *root;
*root = new_node;
}
Node *reverse (Node *root)
{
Node *new_root = NULL;
Node *next;
while ((next = pop(&root))) {
push (&new_root, next);
}
return (new_root);
}
To swap two variables without the use of a temporary variable,
a = a xor b
b = a xor b
a = a xor b
fastest way is to write it in one line
a = a ^ b ^ (b=a)
Similarly,
using two swaps
swap(a,b)
swap(b,c)
solution using xor
a = a^b^c
b = a^b^c
c = a^b^c
a = a^b^c
solution in one line
c = a ^ b ^ c ^ (a=b) ^ (b=c)
b = a ^ b ^ c ^ (c=a) ^ (a=b)
a = a ^ b ^ c ^ (b=c) ^ (c=a)
The same logic is used to reverse a linked list.
typedef struct List
{
int info;
struct List *next;
}List;
List* reverseList(List *head)
{
p=head;
q=p->next;
p->next=NULL;
while(q)
{
q = (List*) ((int)p ^ (int)q ^ (int)q->next ^ (int)(q->next=p) ^ (int)(p=q));
}
head = p;
return head;
}
Here's a simpler version in Java. It does use only two pointers curr & prev
public void reverse(Node head) {
Node curr = head, prev = null;
while (head.next != null) {
head = head.next; // move the head to next node
curr.next = prev; //break the link to the next node and assign it to previous
prev = curr; // we are done with previous, move it to next node
curr = head; // current moves along with head
}
head.next = prev; //for last node
}
Work out the time complexity of the algorithm you are using now and it should be obvious that it can not be improved.
I don't understand why there is need to return head as we are passing it as argument. We are passing head of the link list then we can update also. Below is simple solution.
#include<stdio.h>
#include<conio.h>
struct NODE
{
struct NODE *next;
int value;
};
typedef struct NODE node;
void reverse(node **head);
void add_end(node **head,int val);
void alloc(node **p);
void print_all(node *head);
void main()
{
node *head;
clrscr();
head = NULL;
add_end( &head, 1 );
add_end( &head, 2 );
add_end( &head, 3 );
print_all( head );
reverse( &head );
print_all( head );
getch();
}
void alloc(node **p)
{
node *temp;
temp = (node *) malloc( sizeof(node *) );
temp->next = NULL;
*p = temp;
}
void add_end(node **head,int val)
{
node *temp,*new_node;
alloc(&new_node);
new_node->value = val;
if( *head == NULL )
{
*head = new_node;
return;
}
for(temp = *head;temp->next!=NULL;temp=temp->next);
temp->next = new_node;
}
void print_all(node *head)
{
node *temp;
int index=0;
printf ("\n\n");
if (head == NULL)
{
printf (" List is Empty \n");
return;
}
for (temp=head; temp != NULL; temp=temp->next,index++)
printf (" %d ==> %d \n",index,temp->value);
}
void reverse(node **head)
{
node *next,*new_head;
new_head=NULL;
while(*head != NULL)
{
next = (*head)->next;
(*head)->next = new_head;
new_head = (*head);
(*head) = next;
}
(*head)=new_head;
}
#include <stdio.h>
#include <malloc.h>
tydef struct node
{
int info;
struct node *link;
} *start;
void main()
{
rev();
}
void rev()
{
struct node *p = start, *q = NULL, *r;
while (p != NULL)
{
r = q;
q = p;
p = p->link;
q->link = r;
}
start = q;
}
curr = head;
prev = NULL;
while (curr != NULL) {
next = curr->next; // store current's next, since it will be overwritten
curr->next = prev;
prev = curr;
curr = next;
}
head = prev; // update head
No, nothing faster than the current O(n) can be done. You need to alter every node, so time will be proportional to the number of elements anyway and that's O(n) you already have.
Using two pointers while maintaining time complexity of O(n), the fastest achievable, might only be possible through number casting of pointers and swapping their values. Here is an implementation:
#include <stdio.h>
typedef struct node
{
int num;
struct node* next;
}node;
void reverse(node* head)
{
node* ptr;
if(!head || !head->next || !head->next->next) return;
ptr = head->next->next;
head->next->next = NULL;
while(ptr)
{
/* Swap head->next and ptr. */
head->next = (unsigned)(ptr =\
(unsigned)ptr ^ (unsigned)(head->next =\
(unsigned)head->next ^ (unsigned)ptr)) ^ (unsigned)head->next;
/* Swap head->next->next and ptr. */
head->next->next = (unsigned)(ptr =\
(unsigned)ptr ^ (unsigned)(head->next->next =\
(unsigned)head->next->next ^ (unsigned)ptr)) ^ (unsigned)head->next->next;
}
}
void add_end(node* ptr, int n)
{
while(ptr->next) ptr = ptr->next;
ptr->next = malloc(sizeof(node));
ptr->next->num = n;
ptr->next->next = NULL;
}
void print(node* ptr)
{
while(ptr = ptr->next) printf("%d ", ptr->num);
putchar('\n');
}
void erase(node* ptr)
{
node *end;
while(ptr->next)
{
if(ptr->next->next) ptr = ptr->next;
else
{
end = ptr->next;
ptr->next = NULL;
free(end);
}
}
}
void main()
{
int i, n = 5;
node* dummy_head;
dummy_head->next = NULL;
for(i = 1; i <= n ; ++i) add_end(dummy_head, i);
print(dummy_head);
reverse(dummy_head);
print(dummy_head);
erase(dummy_head);
}
I have a slightly different approach. I wanted to make use of the existing functions (like insert_at(index), delete_from(index)) to reverse the list (something like a right shift operation). The complexity is still O(n) but the advantage is more reused code. Have a look at another_reverse() method and let me know what you all think.
#include <stdio.h>
#include <stdlib.h>
struct node {
int data;
struct node* next;
};
struct node* head = NULL;
void printList(char* msg) {
struct node* current = head;
printf("\n%s\n", msg);
while (current != NULL) {
printf("%d ", current->data);
current = current->next;
}
}
void insert_beginning(int data) {
struct node* newNode = (struct node*) malloc(sizeof(struct node));
newNode->data = data;
newNode->next = NULL;
if (head == NULL)
{
head = newNode;
} else {
newNode->next = head;
head = newNode;
}
}
void insert_at(int data, int location) {
struct node* newNode = (struct node*) malloc(sizeof(struct node));
newNode->data = data;
newNode->next = NULL;
if (head == NULL)
{
head = newNode;
}
else {
struct node* currentNode = head;
int index = 0;
while (currentNode != NULL && index < (location - 1)) {
currentNode = currentNode->next;
index++;
}
if (currentNode != NULL)
{
if (location == 0) {
newNode->next = currentNode;
head = newNode;
} else {
newNode->next = currentNode->next;
currentNode->next = newNode;
}
}
}
}
int delete_from(int location) {
int retValue = -1;
if (location < 0 || head == NULL)
{
printf("\nList is empty or invalid index");
return -1;
} else {
struct node* currentNode = head;
int index = 0;
while (currentNode != NULL && index < (location - 1)) {
currentNode = currentNode->next;
index++;
}
if (currentNode != NULL)
{
// we've reached the node just one prior to the one we want to delete
if (location == 0) {
if (currentNode->next == NULL)
{
// this is the only node in the list
retValue = currentNode->data;
free(currentNode);
head = NULL;
} else {
// the next node should take its place
struct node* nextNode = currentNode->next;
head = nextNode;
retValue = currentNode->data;
free(currentNode);
}
} // if (location == 0)
else {
// the next node should take its place
struct node* nextNode = currentNode->next;
currentNode->next = nextNode->next;
if (nextNode != NULL
) {
retValue = nextNode->data;
free(nextNode);
}
}
} else {
printf("\nInvalid index");
return -1;
}
}
return retValue;
}
void another_reverse() {
if (head == NULL)
{
printf("\nList is empty\n");
return;
} else {
// get the tail pointer
struct node* tailNode = head;
int index = 0, counter = 0;
while (tailNode->next != NULL) {
tailNode = tailNode->next;
index++;
}
// now tailNode points to the last node
while (counter != index) {
int data = delete_from(index);
insert_at(data, counter);
counter++;
}
}
}
int main(int argc, char** argv) {
insert_beginning(4);
insert_beginning(3);
insert_beginning(2);
insert_beginning(1);
insert_beginning(0);
/* insert_at(5, 0);
insert_at(4, 1);
insert_at(3, 2);
insert_at(1, 1);*/
printList("Original List\0");
//reverse_list();
another_reverse();
printList("Reversed List\0");
/* delete_from(2);
delete_from(2);*/
//printList();
return 0;
}
using 2-pointers....bit large but simple and efficient
void reverse()
{
int n=0;
node *temp,*temp1;
temp=strptr;
while(temp->next!=NULL)
{
n++; //counting no. of nodes
temp=temp->next;
}
// we will exchange ist by last.....2nd by 2nd last so.on....
int i=n/2;
temp=strptr;
for(int j=1;j<=(n-i+1);j++)
temp=temp->next;
// i started exchanging from in between ....so we do no have to traverse list so far //again and again for exchanging
while(i>0)
{
temp1=strptr;
for(int j=1;j<=i;j++)//this loop for traversing nodes before n/2
temp1=temp1->next;
int t;
t=temp1->info;
temp1->info=temp->info;
temp->info=t;
i--;
temp=temp->next;
//at the end after exchanging say 2 and 4 in a 5 node list....temp will be at 5 and we will traverse temp1 to ist node and exchange ....
}
}
#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
struct node
{
int data;
struct node *link;
};
struct node *first=NULL,*last=NULL,*next,*pre,*cur,*temp;
void create()
{
cur=(struct node*) malloc(sizeof(struct node));
printf("enter first data to insert");
scanf("%d",&cur->data);
first=last=cur;
first->link=NULL;
}
void insert()
{
int pos,c;
cur=(struct node*) malloc(sizeof(struct node));
printf("enter data to insert and also its position");
scanf("%d%d",&cur->data,&pos);
if(pos==1)
{
cur->link=first;
first=cur;
}
else
{
c=1;
next=first;
while(c<pos)
{
pre=next;
next=next->link;
c++;
}
if(pre==NULL)
{
printf("Invalid position");
}
else
{
cur->link=pre->link;
pre->link=cur;
}
}
}
void display()
{
cur=first;
while(cur!=NULL)
{
printf("data= %d\t address= %u\n",cur->data,cur);
cur=cur->link;
}
printf("\n");
}
void rev()
{
pre=NULL;
cur=first;
while(cur!=NULL)
{
next=cur->link;
cur->link=pre;
pre=cur;
cur=next;
}
first=pre;
}
void main()
{
int choice;
clrscr();
do
{
printf("Options are: -\n1:Create\n2:Insert\n3:Display\n4:Reverse\n0:Exit\n");
printf("Enter your choice: - ");
scanf("%d",&choice);
switch(choice)
{
case 1:
create();
break;
case 2:
insert();
break;
case 3:
display();
break;
case 4:
rev();
break;
case 0:
exit(0);
default:
printf("wrong choice");
}
}
while(1);
}
Yes there is a way using only two pointers. That is by creating new linked list where the first node is the first node of the given list and second node of the first list is added at the start of the new list and so on.
Here is my version:
void reverse(ListElem *&head)
{
ListElem* temp;
ListElem* elem = head->next();
ListElem* prev = head;
head->next(0);
while(temp = elem->next())
{
elem->next(prev);
prev = elem;
elem = temp;
}
elem->next(prev);
head = elem;
}
where
class ListElem{
public:
ListElem(int val): _val(val){}
ListElem *next() const { return _next; }
void next(ListElem *elem) { _next = elem; }
void val(int val){ _val = val; }
int val() const { return _val;}
private:
ListElem *_next;
int _val;
};
I am using java to implement this and approach is test driven development hence test cases are also attached.
The Node class that represent single node -
package com.adnan.linkedlist;
/**
* User : Adnan
* Email : sendtoadnan#gmail.com
* Date : 9/21/13
* Time : 12:02 PM
*/
public class Node {
public Node(int value, Node node){
this.value = value;
this.node = node;
}
private int value;
private Node node;
public int getValue() {
return value;
}
public Node getNode() {
return node;
}
public void setNode(Node node){
this.node = node;
}
}
Service class that takes start node as input and reserve it without using extra space.
package com.adnan.linkedlist;
/**
* User : Adnan
* Email : sendtoadnan#gmail.com
* Date : 9/21/13
* Time : 11:54 AM
*/
public class SinglyLinkedListReversal {
private static final SinglyLinkedListReversal service
= new SinglyLinkedListReversal();
public static SinglyLinkedListReversal getService(){
return service;
}
public Node reverse(Node start){
if (hasOnlyNodeInLinkedList(start)){
return start;
}
Node firstNode, secondNode, thirdNode;
firstNode = start;
secondNode = firstNode.getNode();
while (secondNode != null ){
thirdNode = secondNode.getNode();
secondNode.setNode(firstNode);
firstNode = secondNode;
secondNode = thirdNode;
}
start.setNode(null);
return firstNode;
}
private boolean hasOnlyNodeInLinkedList(Node start) {
return start.getNode() == null;
}
}
And The test case that covers above scenario. Please note that you require junit jars. I am using testng.jar; you can use any whatever pleases you..
package com.adnan.linkedlist;
import org.testng.annotations.Test;
import static org.testng.AssertJUnit.assertTrue;
/**
* User : Adnan
* Email : sendtoadnan#gmail.com
* Date : 9/21/13
* Time : 12:11 PM
*/
public class SinglyLinkedListReversalTest {
private SinglyLinkedListReversal reversalService =
SinglyLinkedListReversal.getService();
#Test
public void test_reverseSingleElement() throws Exception {
Node node = new Node(1, null);
reversalService.reverse(node);
assertTrue(node.getNode() == null);
assertTrue(node.getValue() == 1);
}
//original - Node1(1) -> Node2(2) -> Node3(3)
//reverse - Node3(3) -> Node2(2) -> Node1(1)
#Test
public void test_reverseThreeElement() throws Exception {
Node node3 = new Node(3, null);
Node node2 = new Node(2, node3);
Node start = new Node(1, node2);
start = reversalService.reverse(start);
Node test = start;
for (int i = 3; i >=1 ; i -- ){
assertTrue(test.getValue() == i);
test = test.getNode();
}
}
#Test
public void test_reverseFourElement() throws Exception {
Node node4 = new Node(4, null);
Node node3 = new Node(3, node4);
Node node2 = new Node(2, node3);
Node start = new Node(1, node2);
start = reversalService.reverse(start);
Node test = start;
for (int i = 4; i >=1 ; i -- ){
assertTrue(test.getValue() == i);
test = test.getNode();
}
}
#Test
public void test_reverse10Element() throws Exception {
Node node10 = new Node(10, null);
Node node9 = new Node(9, node10);
Node node8 = new Node(8, node9);
Node node7 = new Node(7, node8);
Node node6 = new Node(6, node7);
Node node5 = new Node(5, node6);
Node node4 = new Node(4, node5);
Node node3 = new Node(3, node4);
Node node2 = new Node(2, node3);
Node start = new Node(1, node2);
start = reversalService.reverse(start);
Node test = start;
for (int i = 10; i >=1 ; i -- ){
assertTrue(test.getValue() == i);
test = test.getNode();
}
}
#Test
public void test_reverseTwoElement() throws Exception {
Node node2 = new Node(2, null);
Node start = new Node(1, node2);
start = reversalService.reverse(start);
Node test = start;
for (int i = 2; i >=1 ; i -- ){
assertTrue(test.getValue() == i);
test = test.getNode();
}
}
}
A simple algorithm if you use the linked list as a stack structure:
#include <stdio.h>
#include <stdlib.h>
typedef struct list {
int key;
char value;
struct list* next;
} list;
void print(list*);
void add(list**, int, char);
void reverse(list**);
void deleteList(list*);
int main(void) {
list* head = NULL;
int i=0;
while ( i++ < 26 ) add(&head, i, i+'a');
printf("Before reverse: \n");
print(head);
printf("After reverse: \n");
reverse(&head);
print(head);
deleteList(head);
}
void deleteList(list* l) {
list* t = l;
while ( t != NULL ) {
list* tmp = t;
t = t->next;
free(tmp);
}
}
void print(list* l) {
list* t = l;
while ( t != NULL) {
printf("%d:%c\n", t->key, t->value);
t = t->next;
}
}
void reverse(list** head) {
list* tmp = *head;
list* reversed = NULL;
while ( tmp != NULL ) {
add(&reversed, tmp->key, tmp->value);
tmp = tmp->next;
}
deleteList(*head);
*head = reversed;
}
void add(list** head, int k, char v) {
list* t = calloc(1, sizeof(list));
t->key = k; t->value = v;
t->next = *head;
*head = t;
}
The performance may be affected since additional function call to the add and malloc so the algorithms of address swaps are better but that one actually creates new list so you can use additional options like sort or remove items if you add a callback function as parameter to the reverse.
Here is a slightly different, but simple approach in C++11:
#include <iostream>
struct Node{
Node(): next(NULL){}
Node *next;
std::string data;
};
void printlist(Node* l){
while(l){
std::cout<<l->data<<std::endl;
l = l->next;
}
std::cout<<"----"<<std::endl;
}
void reverse(Node*& l)
{
Node* prev = NULL;
while(l){
auto next = l->next;
l->next = prev;
prev=l;
l=next;
}
l = prev;
}
int main() {
Node s,t,u,v;
s.data = "1";
t.data = "2";
u.data = "3";
v.data = "4";
s.next = &t;
t.next = &u;
u.next = &v;
Node* ptr = &s;
printlist(ptr);
reverse(ptr);
printlist(ptr);
return 0;
}
Output here
Following is one implementation using 2 pointers (head and r)
ListNode * reverse(ListNode* head) {
ListNode *r = NULL;
if(head) {
r = head->next;
head->next = NULL;
}
while(r) {
head = reinterpret_cast<ListNode*>(size_t(head) ^ size_t(r->next));
r->next = reinterpret_cast<ListNode*>(size_t(r->next) ^ size_t(head));
head = reinterpret_cast<ListNode*>(size_t(head) ^ size_t(r->next));
head = reinterpret_cast<ListNode*>(size_t(head) ^ size_t(r));
r = reinterpret_cast<ListNode*>(size_t(r) ^ size_t(head));
head = reinterpret_cast<ListNode*>(size_t(head) ^ size_t(r));
}
return head;
}
here is a little simple solution...
void reverse()
{
node * pointer1 = head->next;
if(pointer1 != NULL)
{
node *pointer2 = pointer1->next;
pointer1->next = head;
head->next = NULL;
head = pointer1;
if(pointer2 != NULL)
{
while(pointer2 != NULL)
{
pointer1 = pointer2;
pointer2 = pointer2->next;
pointer1->next = head;
head = pointer1;
}
pointer1->next = head;
head = pointer1;
}
}
}
You can have solution of this problem with help of only one extra pointer, that has to be static for the reverse function. It's in O(n) complexity.
#include<stdio.h>
#include<stdlib.h>
typedef struct List* List;
struct List {
int val;
List next;
};
List reverse(List list) { /* with recursion and one static variable*/
static List tail;
if(!list || !list->next) {
tail = list;
return tail;
} else {
reverse1(list->next);
list->next->next = list;
list->next = NULL;
return tail;
}
}
As an alternative, you can use recursion-
struct node* reverseList(struct node *head)
{
if(head == NULL) return NULL;
if(head->next == NULL) return head;
struct node* second = head->next;
head->next = NULL;
struct node* remaining = reverseList(second);
second->next = head;
return remaining;
}
class Node {
Node next;
int data;
Node(int item) {
data = item;
next = null;
}
}
public class LinkedList {
static Node head;
//Print LinkedList
public static void printList(Node node){
while(node!=null){
System.out.print(node.data+" ");
node = node.next;
}
System.out.println();
}
//Reverse the LinkedList Utility
public static Node reverse(Node node){
Node new_node = null;
while(node!=null){
Node next = node.next;
node.next = new_node;
new_node = node;
node = next;
}
return new_node;
}
public static void main(String[] args) {
//Creating LinkedList
LinkedList.head = new Node(1);
LinkedList.head.next = new Node(2);
LinkedList.head.next.next = new Node(3);
LinkedList.head.next.next.next = new Node(4);
LinkedList.printList(LinkedList.head);
Node node = LinkedList.reverse(LinkedList.head);
LinkedList.printList(node);
}
}