Question

239. 滑动窗口最大值

English Version

题目描述

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] <= 104
• 1 <= k <= nums.length

解法

方法一：优先队列（大根堆）

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 $k-1$ 个元素加入优先队列，接下来从第 $k$ 个元素开始，将新元素加入优先队列，同时判断堆顶元素是否滑出窗口，如果滑出窗口则将堆顶元素弹出。然后我们将堆顶元素加入结果数组。

时间复杂度 $O(n \times \log k)$，空间复杂度 $O(k)$。其中 $n$ 为数组长度。

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 ans

class 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;
  }
}

方法二：单调队列

这道题也可以使用单调队列来解决。时间复杂度 $O(n)$，空间复杂度 $O(k)$。

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 ans

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, 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
}