原题链接在这里：https://leetcode.com/problems/sliding-window-maximum/
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:
- How about using a data structure such as deque (double-ended queue)?
- The queue size need not be the same as the window’s size.
- Remove redundant elements and the queue should store only elements that need to be considered.
题解：
用deque, 里面存index.
从尾部添加index前先检查deque的尾部index对应的元素nums[deque.getLast()]是否比要添加的元素nums[i]小或者相等，若是，就把尾部index remove掉，一直remove直到遇到比nums[i]大的数或者LinkedList 为空。e.g. 当添加nums[1] = 3的index 1时，最大的数肯定是3，1就没有用了。也就是说如果出现比先添加的数大的数时，先添加的就没有用了。
如此deque里面保存的就是[第一大index, 第二大index, 第三大index, 第四大index...].
若是i - 头部的index >= k, 就说明现在的window大小已经大于了k, 就需要从头remove一次.
当 i+1>=k 是开始记录res. res的坐标为i-k+1, 取ls的头index, 也就是当前窗口的最大index. 把对应的元素加大res中。
Time Complexity: O(n). 每个元素最多进deque一次, 出deque一次. Space O(k).
AC Java:
1 public class Solution { 2 public int[] maxSlidingWindow(int[] nums, int k) { 3 if(k == 0){ 4 return new int[0]; 5 } 6 7 int [] res = new int[nums.length-k+1]; 8 LinkedList<Integer> deque = new LinkedList<Integer>(); 9 for(int i = 0; i<nums.length; i++){ 10 while(!deque.isEmpty() && nums[deque.getLast()]<=nums[i]){ 11 deque.removeLast(); 12 } 13 deque.addLast(i); 14 if(i - deque.getFirst() >= k){ 15 deque.removeFirst(); 16 } 17 if(i+1>=k){ 18 res[i+1-k] = nums[deque.getFirst()]; 19 } 20 } 21 return res; 22 } 23 }