Sliding Window Maximum

题目描述：

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.
解题思路：
对于第一个窗口，即窗口元素为nums[0,k-1]遍历得到最大的元素，记为cur_max，然后把窗口向前滑动一个位置，即窗口元素为nums[1,k]，对于新得到的窗口有三种情况：

（1）nums[k]>=cur_max，直接把cur_max更新为nums[k]即可

（2）nums[k]<cur_max 且nums[0]=cur_max，即新进来的元素小于上一个窗口的最大值，但滑出窗口的元素等于上一个窗口的最大值，这个时候需要遍历新得到的窗口得到新窗口的最大值

（3）nums[k]<cur_max且nums[0]<cur_max，新窗口的最大值cur_max保持不变

AC代码如下：

class Solution {
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
vector<int> ans;
int n = nums.size();
if (n == 0 || k <= 0 || n < k) return ans;
int cur_max = INT_MIN;
for (int i = 0; i < k; ++i){
if (nums[i]>cur_max)
cur_max = nums[i];
}
ans.push_back(cur_max);
for (int i = k; i < n; ++i){
if (nums[i] >= cur_max){
cur_max = nums[i];
} else if(nums[i-k]==cur_max){
cur_max = INT_MIN;
for (int j = i - k + 1; j <= i; ++j){
if (nums[j]>cur_max)
cur_max = nums[j];
}
}
ans.push_back(cur_max);
}
return ans;
}
};