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] <= 1041 <= k <= nums.lengthstruct 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);
}