LeetCode239. 滑动窗口最大值🌟🌟🌟🌟困难

课后作业

问题描述

原文链接:239. 滑动窗口最大值

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

代码实现

Java

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        Deque<Integer> deque = new LinkedList<>();
        int[] res = new int[nums.length - k + 1];
        int index =  0;

        for(int i = 0 ; i < nums.length; i++){
            while(!deque.isEmpty() && nums[deque.peekLast()] <= nums[i]){
                deque.pollLast();
            }

            deque.add(i);

            if(i - deque.peek() >= k){
                deque.poll();
            }

            if(i + 1 >= k){
                res[index++] = nums[deque.peek()];
            }
        }

        return res;
    }
}

Python

class Solution(object):
    def maxSlidingWindow(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: List[int]
        """
        deque = collections.deque()
        res = []

        for i in range(len(nums)):
            while deque and nums[deque[-1]] <= nums[i]:
                deque.pop()

            deque.append(i)

            if deque[0] == i-k:
                deque.popleft()

            if i >= k-1:
                res.append(nums[deque[0]])

        return res

C++

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        deque<int> q;
        vector<int> res;

        for(int i = 0; i < nums.size(); i++){
            while(!q.empty() && nums[q.back()] <= nums[i]){
                q.pop_back();
            }

            q.push_back(i);

            if(q.front() == i-k){
                q.pop_front();
            }

            if(i >= k-1){
                res.push_back(nums[q.front()]);
            }
        }

        return res;
    }
};

Go

func maxSlidingWindow(nums []int, k int) []int {
    var q []int
    res := []int{}

    for i := 0; i < len(nums); i++ {
        for len(q) > 0 && nums[q[len(q)-1]] <= nums[i] {
            q = q[:len(q)-1]
        }

        q = append(q, i)

        if q[0] == i-k {
            q = q[1:]
        }

        if i >= k-1 {
            res = append(res, nums[q[0]])
        }
    }

    return res
}

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