Leetcode221: 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.

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]

.

这道题给定了一个数组，还给了一个窗口大小k，让我们每次向右滑动一个数字，每次返回窗口内的数字的最大值，而且要求我们代码的时间复杂度为O(n)。提示我们要用双向队列deque来解题，并提示我们窗口中只留下有用的值，没用的全移除掉。果然Hard的题目我就是不会做，网上看到了别人的解法才明白，解法又巧妙有简洁，膜拜啊。大概思路是用双向队列保存数字的下标，遍历整个数组，如果此时队列的首元素是i
- k的话，表示此时窗口向右移了一步，则移除队首元素。然后比较队尾元素和将要进来的值，如果小的话就都移除，然后此时我们把队首元素加入结果中即可，参见代码如下：

class Solution {
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
int n = nums.size();
vector<int>res;
deque<int> q;
for(int i = 0; i < n; i++)
{
if(!q.empty() && q.front() == i-k)  q.pop_front();
while(!q.empty() && nums[q.back()] < nums[i])   q.pop_back();
q.push_back(i);
if(i>=k-1)   res.push_back(nums[q.front()]);
}
return res;
}
};