Sliding Window Maximum

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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?

Analyse: Find the maximum in each window and push it into the result vector.

Runtime: 340ms.

 1 class Solution {
 2 public:
 3     vector<int> maxSlidingWindow(vector<int>& nums, int k) {
 4         vector<int> result;
 5         if(nums.size() == 0) return result;
 6         
 7         int left = 0, right = k - 1;
 8         while(right < nums.size()){
 9             int temp = INT_MIN; //maximum in the current window
10             if(left == right) temp = nums[left];
11             else{
12                 for(int i = left; i <= right; i++)
13                     temp = max(temp, nums[i]);
14             }
15             result.push_back(temp);
16             left++;
17             right++;
18         }
19         return result;
20     }
21 };

Sliding Window Maximum

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原文地址：http://www.cnblogs.com/amazingzoe/p/4799783.html