经典题目：239. 滑动窗口最大值

双端队列

class Solution {
    //利用双端队列
    public int[] maxSlidingWindow(int[] nums, int k) {
        Deque<Integer> deque = new LinkedList<>();
        int n = nums.length;
        int[] res = new int[n-k+1];
        for(int i=0 ; i<k ; ++i){
            while(!deque.isEmpty() && nums[i]>nums[deque.peekFirst()]){
                deque.removeFirst();
            }
            deque.offerFirst(i);//栈顶
        }
        res[0] = nums[deque.peekLast()]; //队尾元素为当前窗口最大值
        //队列保证了元素进入的先后顺序，那么就可以保证队尾的元素下标是从小到大的

        for(int i=k ; i<n ; ++i){
            while(!deque.isEmpty() && deque.peekLast() <= i-k){
                deque.removeLast();//超出滑动窗口下表的元素出队
            }
            while(!deque.isEmpty() && nums[i] > nums[deque.peekFirst()]){
                deque.removeFirst();
            }
            deque.offerFirst(i);
            res[i-k+1] = nums[deque.peekLast()];
        }
        return res;

    }
}

优先队列：

class Solution {
    //优先队列
    public int[] maxSlidingWindow(int[] nums, int k) {
        PriorityQueue<int[]> pq = new PriorityQueue<>(new Comparator<int[]>(){ //默认小顶堆
            public int compare(int[] pair1, int[] pair2){
               return pair1[0] == pair2[0]?pair2[1]-pair1[1]:pair2[0]-pair1[0];  //元素相等二点情况下，下表更大的优先排在更前面（相对也更大些）
            }
        });
        int n = nums.length;
        int[] res = new int[n-k+1];
        for(int i=0 ; i<k ; ++i){
            pq.offer(new int[]{nums[i], i});
        }
        res[0] = pq.peek()[0];
        for(int i=k ; i<n ; ++i){
            pq.offer(new int[]{nums[i], i});
            while(pq.peek()[1] <= i-k){
                pq.poll();
            }
            res[i-k+1] = pq.peek()[0];
        }
        return res;
    }
}