[LeetCode]239. Sliding Window Maximum

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].

思路：使用一个优先队列存当前选中的所有数值，按从大到小排序，滑动窗口移动时，从队列中删除滑动窗口前一位的值，然后加入新进入滑动窗口的值，优先队列中最大的即为结果

public class Solution {    public int[] maxSlidingWindow(int[] nums, int k) {        int[] res=new int[nums.length-k+1];        if(nums.length==0){            int[] a=new int[0];            return a;        }        PriorityQueue<Integer> pq=new PriorityQueue<Integer>(11,new Comparator<Integer>(){@Overridepublic int compare(Integer o1, Integer o2) {return o2-o1;}                });        for(int i=0;i<k;i++){            pq.add(nums[i]);        }        int index=0;        res[index]=pq.peek();        index++;        int start=0;        for(int i=k;i<nums.length;i++,index++,start++){            pq.remove((Integer)nums[start]);            pq.add(nums[i]);            res[index]=pq.peek();        }        return res;    }}