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困难 |
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给你一个整数数组 nums，有一个大小为 k 的滑动窗口从数组的最左侧移动到数组的最右侧。你只可以看到在滑动窗口内的 k 个数字。滑动窗口每次只向右移动一位。
返回 滑动窗口中的最大值 。
示例 1：
输入：nums = [1,3,-1,-3,5,3,6,7], 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
示例 2：
输入：nums = [1], k = 1 输出：[1]
提示：
1 <= nums.length <= 105-104 <= nums[i] <= 1041 <= k <= nums.length
我们可以使用优先队列（大根堆）来维护滑动窗口中的最大值。
先将前
时间复杂度
class Solution:
def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
q = [(-v, i) for i, v in enumerate(nums[: k - 1])]
heapify(q)
ans = []
for i in range(k - 1, len(nums)):
heappush(q, (-nums[i], i))
while q[0][1] <= i - k:
heappop(q)
ans.append(-q[0][0])
return ansclass Solution {
public int[] maxSlidingWindow(int[] nums, int k) {
PriorityQueue<int[]> q
= new PriorityQueue<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : b[0] - a[0]);
int n = nums.length;
for (int i = 0; i < k - 1; ++i) {
q.offer(new int[] {nums[i], i});
}
int[] ans = new int[n - k + 1];
for (int i = k - 1, j = 0; i < n; ++i) {
q.offer(new int[] {nums[i], i});
while (q.peek()[1] <= i - k) {
q.poll();
}
ans[j++] = q.peek()[0];
}
return ans;
}
}class Solution {
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
priority_queue<pair<int, int>> q;
int n = nums.size();
for (int i = 0; i < k - 1; ++i) {
q.push({nums[i], -i});
}
vector<int> ans;
for (int i = k - 1; i < n; ++i) {
q.push({nums[i], -i});
while (-q.top().second <= i - k) {
q.pop();
}
ans.emplace_back(q.top().first);
}
return ans;
}
};func maxSlidingWindow(nums []int, k int) (ans []int) {
q := hp{}
for i, v := range nums[:k-1] {
heap.Push(&q, pair{v, i})
}
for i := k - 1; i < len(nums); i++ {
heap.Push(&q, pair{nums[i], i})
for q[0].i <= i-k {
heap.Pop(&q)
}
ans = append(ans, q[0].v)
}
return
}
type pair struct{ v, i int }
type hp []pair
func (h hp) Len() int { return len(h) }
func (h hp) Less(i, j int) bool {
a, b := h[i], h[j]
return a.v > b.v || (a.v == b.v && i < j)
}
func (h hp) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
func (h *hp) Push(v any) { *h = append(*h, v.(pair)) }
func (h *hp) Pop() any { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }use std::collections::VecDeque;
impl Solution {
#[allow(dead_code)]
pub fn max_sliding_window(nums: Vec<i32>, k: i32) -> Vec<i32> {
// The deque contains the index of `nums`
let mut q: VecDeque<usize> = VecDeque::new();
let mut ans_vec: Vec<i32> = Vec::new();
for i in 0..nums.len() {
// Check the first element of queue, if it's out of bound
if !q.is_empty() && (i as i32) - k + 1 > (*q.front().unwrap() as i32) {
// Pop it out
q.pop_front();
}
// Pop back elements out until either the deque is empty
// Or the back element is greater than the current traversed element
while !q.is_empty() && nums[*q.back().unwrap()] <= nums[i] {
q.pop_back();
}
// Push the current index in queue
q.push_back(i);
// Check if the condition is satisfied
if i >= ((k - 1) as usize) {
ans_vec.push(nums[*q.front().unwrap()]);
}
}
ans_vec
}
}/**
* @param {number[]} nums
* @param {number} k
* @return {number[]}
*/
var maxSlidingWindow = function (nums, k) {
let ans = [];
let q = [];
for (let i = 0; i < nums.length; ++i) {
if (q && i - k + 1 > q[0]) {
q.shift();
}
while (q && nums[q[q.length - 1]] <= nums[i]) {
q.pop();
}
q.push(i);
if (i >= k - 1) {
ans.push(nums[q[0]]);
}
}
return ans;
};using System.Collections.Generic;
public class Solution {
public int[] MaxSlidingWindow(int[] nums, int k) {
if (nums.Length == 0) return new int[0];
var result = new int[nums.Length - k + 1];
var descOrderNums = new LinkedList<int>();
for (var i = 0; i < nums.Length; ++i)
{
if (i >= k && nums[i - k] == descOrderNums.First.Value)
{
descOrderNums.RemoveFirst();
}
while (descOrderNums.Count > 0 && nums[i] > descOrderNums.Last.Value)
{
descOrderNums.RemoveLast();
}
descOrderNums.AddLast(nums[i]);
if (i >= k - 1)
{
result[i - k + 1] = descOrderNums.First.Value;
}
}
return result;
}
}这道题也可以使用单调队列来解决。时间复杂度
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)class Solution:
def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
q = deque()
ans = []
for i, v in enumerate(nums):
if q and i - k + 1 > q[0]:
q.popleft()
while q and nums[q[-1]] <= v:
q.pop()
q.append(i)
if i >= k - 1:
ans.append(nums[q[0]])
return ansclass 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, j = 0; i < n; ++i) {
if (!q.isEmpty() && i - k + 1 > q.peekFirst()) {
q.pollFirst();
}
while (!q.isEmpty() && nums[q.peekLast()] <= nums[i]) {
q.pollLast();
}
q.offer(i);
if (i >= k - 1) {
ans[j++] = nums[q.peekFirst()];
}
}
return ans;
}
}class Solution {
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
deque<int> q;
vector<int> ans;
for (int i = 0; i < nums.size(); ++i) {
if (!q.empty() && i - k + 1 > q.front()) {
q.pop_front();
}
while (!q.empty() && nums[q.back()] <= nums[i]) {
q.pop_back();
}
q.push_back(i);
if (i >= k - 1) {
ans.emplace_back(nums[q.front()]);
}
}
return ans;
}
};func maxSlidingWindow(nums []int, k int) (ans []int) {
q := []int{}
for i, v := range nums {
if len(q) > 0 && i-k+1 > q[0] {
q = q[1:]
}
for len(q) > 0 && nums[q[len(q)-1]] <= v {
q = q[:len(q)-1]
}
q = append(q, i)
if i >= k-1 {
ans = append(ans, nums[q[0]])
}
}
return ans
}