[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>(){
			@Override
			public 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;
    }
}