[LeetCode] 239. Sliding Window Maximum

Problem

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. Return the max sliding window.

Example:

Input: nums = [1,3,-1,-3,5,3,6,7], and k = 3
Output: [3,3,5,5,6,7] 
Explanation: 

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

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

Follow up:
Could you solve it in linear time?

Solution

using max heap O(nlogn)

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        PriorityQueue<Integer> queue = new PriorityQueue<>((i1, i2)->i2-i1);
        if (nums == null || k == 0 || nums.length < k) return new int[0];
        int[] res = new int[nums.length-k+1];
        for (int i = 0; i < k; i++) queue.offer(nums[i]);
        res[0] = queue.peek();
        for (int i = k; i < nums.length; i++) {
            queue.remove(nums[i-k]);
            queue.offer(nums[i]);
            res[i-k+1] = queue.peek();
        }
        return res;
    }
}

using Deque to maintain a window O(n)

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        if (nums == null || k == 0 || nums.length < k) return new int[0];
        int[] res = new int[nums.length-k+1];
        int index = 0;
        Deque<Integer> queue = new ArrayDeque<>(); //queue to save index
        for (int i = 0; i < nums.length; i++) {
            //丢弃队首那些超出窗口长度的元素 index < i-k+1
            if (!queue.isEmpty() && queue.peek() < i-k+1) queue.poll();
            //队首的元素都是比后来加入元素大的元素，所以存储的index对应的元素是从小到大
            while (!queue.isEmpty() && nums[queue.peekLast()] < nums[i]) queue.pollLast();
            queue.offer(i);
            if (i >= k-1) res[index++] = nums[queue.peek()];
        }
        return res;
    }
}