返回滑动窗口中的最大值 Sliding Window Maximum

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. 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?

解法一：暴力法

//1. 遍历求出每个滑动窗口
//2. 在每个滑动窗口中求出最大值，记录下来。
//3. 注意边界情况的处理，如：输入的数组为null，或者为空；滑动窗口的大小为零，或者大于数组长度等。

import java.util.ArrayList;

public class Solution {
public ArrayList<Integer> maxInWindows(int [] num, int size)
{
ArrayList<Integer> arr = new ArrayList<Integer>();
if(num==null || num.length==0 || size==0){
return arr;
}
int len=num.length, inx=0, end=len-1;
if(size>len){
return arr;
}else if(size==len){
arr.add(maxNum(num, 0, len-1));
}else{
for(int i=0; i<len; i++){
inx = i;
end = inx+size-1;
if(end >= len){
break;
}else{
arr.add(maxNum(num, inx, end));
}
}
}
return arr;
}

//求几个值中的最大值
public int maxNum(int[] num, int inx, int end){
int max = num[inx];
for(int i=inx; i<=end; i++){
if(num[i]>max){
max = num[i];
}
}
return max;
}
}

解法二：双端队列大顶堆简化版 Max Heap

//1. 滑动窗口应当是队列，但为了得到滑动窗口的最大值，队列序可以从两端删除元素，因此使用双端队列。
//2. 对新来的元素k，将其与双端队列中的元素相比较, 前面比k小的，直接移出队列（因为不再可能成为后面滑动窗口的最大值了!
//3. 前面比k大的X，比较两者下标，判断X是否已不在窗口之内，不在了，直接移出队列。队列的第一个元素是当前滑动窗口中的最大值
public ArrayList<Integer> maxInWindows(int [] num, int size)
    {
        ArrayList<Integer> res = new ArrayList<>();
        if(size == 0) return res;
        int begin;
        ArrayDeque<Integer> q = new ArrayDeque<>();
        for(int i = 0; i < num.length; i++){
            begin = i - size + 1;
            if(q.isEmpty())
                q.add(i);
            else if(begin > q.peekFirst())
                q.pollFirst();

            while((!q.isEmpty()) && num[q.peekLast()] <= num[i])
                q.pollLast();
            q.add(i);
            if(begin >= 0)
                res.add(num[q.peekFirst()]);
        }
        return res;
    }

返回滑动窗口中的最大值 Sliding Window Maximum

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