239. Sliding Window Maximum

难度: Hard

题目描述

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

解题思路

代码实现

解决方案

java

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {

        int n = nums.length;
        Deque<Integer> dq = new LinkedList<>();

        for (int i = 0; i < k; i++) {
            while (!dq.isEmpty() && nums[i] >= nums[dq.peekLast()]) {
                dq.pollLast();
            }
            dq.offerLast(i);

        }

        int[] ans = new int[n - k + 1];
        ans[0] = nums[dq.peekFirst()];
        for (int i = k; i < n; i++) {
            while (!dq.isEmpty() && nums[i] >= nums[dq.peekLast()]) {
                dq.pollLast();
            }
            dq.offerLast(i);

            while (dq.peekFirst() <= i - k) {
                dq.pollFirst();
            }
            ans[i - k + 1] = nums[dq.peekFirst()];

        }

        return ans;

    }
}