[LeetCode] 239. Sliding Window Maximum

lentoo 发布于2019-08-16 13:45 / 339人阅读

摘要：丢弃队首那些超出窗口长度的元素队首的元素都是比后来加入元素大的元素，所以存储的对应的元素是从小到大

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 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 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;
    }
}

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