题目描述
给定一个数组 nums 和滑动窗口的大小 k，请找出所有滑动窗口里的最大值。
示例:
输入: nums = [1,3,-1,-3,5,3,6,7], 和 k = 3
输出: [3,3,5,5,6,7]
解释:
滑动窗口的位置 最大值
--------------- -----
[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
提示：
你可以假设 k 总是有效的，在输入数组不为空的情况下，1 ≤ k ≤ 输入数组的大小。
注意：本题与主站 239 题相同：https://leetcode.cn/problems/sliding-window-maximum/
解法
方法一：单调队列
单调队列常见模型：找出滑动窗口中的最大值/最小值。模板：
q = deque()
for i in range(n):
# 判断队头是否滑出窗口
while q and checkout_out(q[0]):
q.popleft()
while q and check(q[-1]):
q.pop()
q.append(i)
时间复杂度 $O(n)$，空间复杂度 $O(k)$。其中 $n$ 为数组长度。
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13 | class Solution:
def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
q = deque()
ans = []
for i, x in enumerate(nums):
if q and i - q[0] + 1 > k:
q.popleft()
while q and nums[q[-1]] <= x:
q.pop()
q.append(i)
if i >= k - 1:
ans.append(nums[q[0]])
return ans
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20 | class Solution {
public int[] maxSlidingWindow(int[] nums, int k) {
int n = nums.length;
int[] ans = new int[n - k + 1];
Deque<Integer> q = new ArrayDeque<>();
for (int i = 0; i < n; ++i) {
if (!q.isEmpty() && i - q.peek() + 1 > k) {
q.poll();
}
while (!q.isEmpty() && nums[q.peekLast()] <= nums[i]) {
q.pollLast();
}
q.offer(i);
if (i >= k - 1) {
ans[i - k + 1] = nums[q.peek()];
}
}
return ans;
}
}
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21 | class Solution {
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
vector<int> ans;
deque<int> q;
int n = nums.size();
for (int i = 0; i < n; ++i) {
if (!q.empty() && i - q.front() + 1 > k) {
q.pop_front();
}
while (!q.empty() && nums[q.back()] <= nums[i]) {
q.pop_back();
}
q.push_back(i);
if (i >= k - 1) {
ans.push_back(nums[q.front()]);
}
}
return ans;
}
};
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16 | func maxSlidingWindow(nums []int, k int) (ans []int) {
q := []int{}
for i, x := range nums {
for len(q) > 0 && i-q[0]+1 > k {
q = q[1:]
}
for len(q) > 0 && nums[q[len(q)-1]] <= x {
q = q[:len(q)-1]
}
q = append(q, i)
if i >= k-1 {
ans = append(ans, nums[q[0]])
}
}
return
}
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18 | function maxSlidingWindow(nums: number[], k: number): number[] {
const q: number[] = [];
const n = nums.length;
const ans: number[] = [];
for (let i = 0; i < n; ++i) {
while (q.length && i - q[0] + 1 > k) {
q.shift();
}
while (q.length && nums[q[q.length - 1]] <= nums[i]) {
q.pop();
}
q.push(i);
if (i >= k - 1) {
ans.push(nums[q[0]]);
}
}
return ans;
}
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22 | use std::collections::VecDeque;
impl Solution {
pub fn max_sliding_window(nums: Vec<i32>, k: i32) -> Vec<i32> {
let k = k as usize;
let n = nums.len();
let mut ans = vec![0; n - k + 1];
let mut q = VecDeque::new();
for i in 0..n {
while !q.is_empty() && i - q[0] + 1 > k {
q.pop_front();
}
while !q.is_empty() && nums[*q.back().unwrap()] <= nums[i] {
q.pop_back();
}
q.push_back(i);
if i >= k - 1 {
ans[i - k + 1] = nums[q[0]];
}
}
ans
}
}
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23 | /**
* @param {number[]} nums
* @param {number} k
* @return {number[]}
*/
var maxSlidingWindow = function (nums, k) {
const q = [];
const n = nums.length;
const ans = [];
for (let i = 0; i < n; ++i) {
while (q.length && i - q[0] + 1 > k) {
q.shift();
}
while (q.length && nums[q[q.length - 1]] <= nums[i]) {
q.pop();
}
q.push(i);
if (i >= k - 1) {
ans.push(nums[q[0]]);
}
}
return ans;
};
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19 | public class Solution {
public int[] MaxSlidingWindow(int[] nums, int k) {
if (nums.Length == 0) {
return new int[]{};
}
int[] array = new int[nums.Length - (k - 1)];
Queue<int> queue = new Queue<int>();
int index = 0;
for (int i = 0; i < nums.Length; i++) {
queue.Enqueue(nums[i]);
if (queue.Count == k) {
array[index] = queue.Max();
queue.Dequeue();
index++;
}
}
return array;
}
}
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