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?
```
## 思路
符合直觉的想法是直接遍历 nums, 然后然后用一个变量 slideWindow 去承载 k 个元素，
然后对 slideWindow 求最大值，这是可以的，时间复杂度是 O(n \* k).代码如下：
```js
/**
* @param {number[]} nums
* @param {number} k
* @return {number[]}
*/
varmaxSlidingWindow=function(nums,k){
// bad 时间复杂度O(n * k)
if(nums.length===0||k===0)return[];
letslideWindow=[];
constret=[];
for(leti=0;i<nums.length-k+1;i++){
for(letj=0;j<k;j++){
slideWindow.push(nums[i+j]);
}
ret.push(Math.max(...slideWindow));
slideWindow=[];
}
returnret;
};
```
但是如果真的是这样，这道题也不会是 hard 吧？这道题有一个 follow up，要求你用线性的时间去完成。
