Leetcode 239 Sliding Window Maximum-优快云博客

Given an array 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.

For example,
Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.

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

Therefore, return the max sliding window as [3,3,5,5,6,7].

滑动窗口的问题。需要留下window里面的最大值。

可以用LinkedList来实现

public class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        if(nums == null || nums.length == 0) {
            return new int[0];
        }
        
        LinkedList<Integer> list = new LinkedList<>();//list里面存index
        int[] result = new int[nums.length - k + 1];
        
        for(int i = 0; i < nums.length; i ++){
            while(!list.isEmpty() && list.peek() < i - k + 1){//把window外面的挪出去
                list.poll();
            }
            
            while(!list.isEmpty() && nums[list.peekLast()] < nums[i]){//把比当前要加的新元素小的挪出去 也就是window里面只留下最大值
                list.removeLast();
            }
            list.offerLast(i);
            if((i + 1) >= k) {//c从i = k-1的地方开始，也就是第一次填满window的地方
                result[i + 1 - k] = nums[list.peek()];
            }
        }
        return result;
    }
}