How to convert iter.next() to an array index? - arrays

Question

I would implement an array v to mark the current vertex as visited. However, iter.next() cannot be an index of LinkedList or an index of an array.
The output of iter.next() is not an integer, it is an option Some.
How to use the value in Some? If I would use value in Some as an index of array. Should I convert the value into an integer?
fn bfs(s: usize) {
let mut cells = vec![LinkedList::<usize>::new(); 4];
for (i, cell) in cells.iter_mut().enumerate() {
cell.push_front(i);
}
cells[0].push_back(1);
cells[0].push_back(2);
cells[1].push_back(2);
cells[2].push_back(0);
cells[2].push_back(3);
cells[3].push_back(3);
let mut iter = cells[s].iter();
let mut v = vec![0; 4];
// Entry point
for i in 0..cells[s].len() {
// It is incorrect, "iter.next()" it cannot be a index of LinkedList
// vector "v" to keep track of which vertex has been traversed
//v[iter.next()] = 1;
}
}
fn main() {
bfs(2);
}

Answer 1

iter.next() return an Option, you just need to deconstruct it:
for i in 0..cells[s].len() {
let x = match iter.next() {
Some(x) => *x,
None => return,
};
v[x] = 1;
}

Answer 2

Rust's for loops can iterate for you, so you don't need to explicitly create an iterator or call next(). The Option will already be unwrapped inside the loop, because the loop automatically ends when next() returns None.
for &i in &cells[s] {
v[i] = 1;
}
This saves you having to keep track of index variables just to create a loop. And the code is a lot shorter and more readable.

Answer 3

This is the final version
fn bfs(s: usize) {
let mut cells = vec![LinkedList::<usize>::new(); 4];
let mut traversal = LinkedList::new();
let mut queue = LinkedList::new();
for (i, cell) in cells.iter_mut().enumerate() {
cell.push_front(i);
}
cells[0].push_back(1);
cells[0].push_back(2);
cells[1].push_back(2);
cells[2].push_back(0);
cells[2].push_back(3);
cells[3].push_back(3);
let mut v = vec![0; 4];
let mut done = false;
let c = s;
while !done {
for &x in &cells[c] {
if v[x] == 0 {
v[x] = 1;
queue.push_back(x);
traversal.push_back(x)
}
}
let c = queue.pop_back();
if queue.is_empty() {
// walkthrough
for i in 0..cells.len() {
if v[i] == 0 {
v[i] = 1;
queue.push_back(i);
traversal.push_back(i)
}
}
if queue.is_empty() {
done = true;
}
}
}
println!("{:?}", traversal);
}

Answer 4

You can use the reference of the integer value in Some as the index of the array.
let mut v = vec![0; 4];
while !done {
let mut iter = cells[c].iter().peekable();
// Entry point
while let Some(&x) = iter.next_if(|&&n| v[n] == 0) {
v[x] = 1;
queue.push_back(x);
traversal.push_back(x)
}
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