239. Sliding Window Maximum

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题目：

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?

Hint:

1. How about using a data structure such as deque (double-ended queue)?
2. The queue size need not be the same as the window’s size.
3. Remove redundant elements and the queue should store only elements that need to be considered.

链接： http://leetcode.com/problems/sliding-window-maximum/

题解：

长题目思密达。题目大意是给定一个数组和一个长为k的window，求这个window在从头到尾滑动时，每个window里的最大值组成的集合。考虑了很久，最后求助了Discuss。原理使用一个Deque，或者用doubly linkedlist也可以。我们队这个数组维护一个递减的双端队列，队列的内容为坐标index。每次移动时，移除小于当前队首坐标的元素，同时比较当前元素与队尾元素的大小，假如队尾元素较小，移除队尾元素，继续比较当前元素和队尾。比较完毕后把当前元素加入到队列中， 最后判断是否窗口已满，要输出到结果集中。

Time Complexity - O(n)， Space Complexity - O(n)。

public class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        if(nums == null || nums.length == 0)
            return new int[]{};
        int len = nums.length;
        int[] res = new int[len - k + 1];
        Deque<Integer> dq = new LinkedList<>();         
        
        for(int i = 0; i < len; i++) {
            while(!dq.isEmpty() && dq.peekFirst() < i - (k - 1))    // maintain a window of length k
                dq.pollFirst();
            
            while(!dq.isEmpty() && nums[dq.peekLast()] < nums[i])  // compare last elements with nums[i]
                dq.pollLast();
                
            dq.offerLast(i);
            
            if(i >= k - 1)
                res[i - (k - 1)] = nums[dq.peekFirst()];    // since we have a descendent deque, first is always the largest in window
        }
        
        return res;
    }
}

题外话：

最近感恩节，祝福大家感恩节快乐。这几天总在网上看deal，没有好好刷题，难得的4天休假也没有完全利用起来。Cousera的斯坦福算法1总算上完了，跟得很累，每周都要花不少时间。但是有一些思想和题目比较经典，比如closest pair，find inversions，matrix multiply，还有find median in data stream等等。有时间要好好复习一下。下面记录几个link有空时学习。

https://leetcode.com/discuss/64811/easy-to-understand-double-heap-solution-in-java

http://andrew-algorithm.blogspot.com/

下周上一周班，周四有公司年会，之后那周是培训5天，然后我打算请两周假，这样整个十二月份基本都能有时间好好做题和学习。 抽空要把加拿大签证办了，然后过去签美国签证，这样明年二月就可以顺利回国和顺便去趟日本玩了。

Reference:

239. Sliding Window Maximum

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原文地址：http://www.cnblogs.com/yrbbest/p/5004596.html