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 = [1], k = 1
Output: [1]

Example 3:

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

Example 4:

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

Example 5:

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

 

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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