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239. Sliding Window Maximum

中文文档

Description

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]

 

Constraints:

  • 1 <= nums.length <= 105
  • -104 <= nums[i] <= 104
  • 1 <= k <= nums.length

Solutions

Python3

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

Java

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;
    }
}

JavaScript

/**
 * @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;
};

C++

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.push_back(nums[q.front()]);
        }
        return ans;
    }
};

Go

func maxSlidingWindow(nums []int, k int) []int {
	var q []int
	var ans []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
}

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