Array to List in C (iteratively) - c

Question

Using the typedefinitions:
typedef struct Node *List;
typedef struct Node {
int item;
List next;
} Node;
The recursive solution I found:
List arrayToList(int arr[],int n,int idx) {
if (n==idx) return NULL;
List list=malloc(sizeof(Node));
list->next=arrayToList(arr,n,idx+1);
list->item=arr[idx];
return list;
}
UPDATE: The following is almost correct, but I don't know why a zero is printed in the end.
List newNode() {
List li=malloc(sizeof(Node));
return li;
}
List arrayToList(int arr[],int n) {
List li=newNode();
List li1=li; /*save the beginning of the list*/
int i;
for (i=0;i<n;i++) {
li->item=arr[i];
li->next=newNode();
li=li->next;
}
li=NULL;
return li1;
}
void printList(List li) {
while (li!=NULL) {
li=li->next;
}
printf("\n");
}
int main(int argc, char* argv[]) {
int arr[]={4,1,2,3,4,7,4,5,6,8};
List li=arrayToList(arr,10);
printList(li);
return 0;
}
The output I get is: 4 1 2 3 4 7 4 5 6 8 0 .
UPDATE 2: Changing the function printList into this one gives me correct output:
void printList(List li) {
while (li->next!=NULL) {
printf("%d ",li->item);
li=li->next;
}
printf("\n");
}
But I'm left wondering why should I have li->next!=NULL as guard?

Answer 1

Do this:
typedef struct node {
int data;
struct node *next;
} Node;
Note that you want next to point to the next node, not list.
Node* arrayToList(int arr[],int n,int idx) {
if (n==idx) return NULL;
Node* headOfList = (Node*) malloc(sizeof(Node));
Node* tail = headOfList;
int i = n;
for(; i < idx; i++) {
// Create node
Node* node = (Node*) malloc(sizeof(Node));
node->data = arr[i];
node->next = NULL;
// Link Node to current list
tail->next = node;
tail = node;
}
return headOfList;
}
This is just a rough draft (I've not tried it out with a compiler), but the idea is there.

Answer 2

Initialize your first element:
Node *head = (Node*) malloc(sizeof(Node));
head->data = arr[0];
You will need a pointer to keep track of your current position:
Node *curr = head;
Create a for loop:
for(int i = 1; i < len; i++)
{
}
Then for each element, add a new node and move your pointer forward.
curr->next = (Node*) malloc(sizeof(Node));
curr = curr->next;
curr->data = arr[i];
Don't forget to set the last element of your list to point to null! (This can happen outside the for loop.)
n->next = NULL;