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].

Note:
You may assume k is always valid, ie: 1 ≤ k ≤ input array’s size for non-empty array.

Follow up:
Could you solve it in linear time?

 堆排，时间复杂度->(nlog(n)) 关健的问题在于…每次调整都是logn的复杂度

class Solution {
    
    public int[] maxSlidingWindow(int[] nums, int k) {
        //init
        if(nums.length == 0 || nums.length < k)return new int[0];
        int[] ret = new int[nums.length - k  + 1];
        PriorityQueue heap = new PriorityQueue();

        //init
        for(int i =0 ; i

 双端队列的思路

class Solution {
    
    Deque queue = new LinkedList<>();

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

        //bound
        if(k == 0 || k > nums.length)return new int[0];
        
        int[] ret = new int[nums.length - k + 1];

        int left = 0;
        int right = 0;
        while(right < nums.length){
            add(nums[right]);
            if(right - left + 1 == k){
                ret[right - k + 1] = queue.peekFirst();
                remove(nums[left++]);
            }
            right++;
        }
        
        return ret;
    }

    public void add(int val){
        while (!queue.isEmpty()){
            if (queue.peekLast() < val){
                queue.pollLast();
            }else{
                queue.addLast(val);
                return;
            }
        }

        queue.addLast(val);
    }

    public void remove(int val){
        if(queue.peekFirst() == val){
            queue.removeFirst();
        }
    }
}