## 239. Sliding Window Maximum

You are given an array of integers `nums`, there is a sliding window of size `k` which is moving from the very left of the array to the very right. You can only see the `k` numbers in the window. Each time the sliding window moves right by one position.

Return the max sliding window.

Example 1:

```Input: nums = [1,3,-1,-3,5,3,6,7], k = 3
Output: [3,3,5,5,6,7]
Explanation:
Window position                Max
---------------               -----
[1  3  -1] -3  5  3  6  7       3
1 [3  -1  -3] 5  3  6  7       3
1  3 [-1  -3  5] 3  6  7       5
1  3  -1 [-3  5  3] 6  7       5
1  3  -1  -3 [5  3  6] 7       6
1  3  -1  -3  5 [3  6  7]      7
```

Example 2:

```Input: nums = , k = 1
Output: 
```

Example 3:

```Input: nums = [1,-1], k = 1
Output: [1,-1]
```

Example 4:

```Input: nums = [9,11], k = 2
Output: 
```

Example 5:

```Input: nums = [4,-2], k = 2
Output: 
```

Constraints:

• `1 <= nums.length <= 105`
• `-104 <= nums[i] <= 104`
• `1 <= k <= nums.length`

## Rust Solution

``````struct Solution;
use std::collections::VecDeque;

impl Solution {
fn max_sliding_window(nums: Vec<i32>, k: i32) -> Vec<i32> {
let k = k as usize;
let n = nums.len();
let mut queue: VecDeque<usize> = VecDeque::new();
let mut res = vec![];
for i in 0..n {
let n = queue.len();
for _ in 0..n {
let j = queue.pop_front().unwrap();
if i - j < k && nums[j] >= nums[i] {
queue.push_back(j);
}
}
queue.push_back(i);
if i + 1 >= k {
res.push(nums[*queue.front().unwrap()]);
}
}
res
}
}

#[test]
fn test() {
let nums = vec![1, 3, -1, -3, 5, 3, 6, 7];
let k = 3;
let res = vec![3, 3, 5, 5, 6, 7];
assert_eq!(Solution::max_sliding_window(nums, k), res);
}
``````

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