Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.

For example, given the following triangle

[     [2],    [3,4],   [6,5,7],  [4,1,8,3]]

The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).

Note:

Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.

//利用dp来解决，更好的办法是从后向前计算，这样就能够使用一维数组空间的来解决，符合题目的意思//那么。动态转移方程为。dp[j] = triangle[i][j] + min(dp[j],dp[j+1])class Solution {public:    int minimumTotal(std::vector
     
      
        > &triangle) {		int n = triangle.size();		if(n < 1) return 0;		std::vector
       
         dp(triangle[n-1]);		for(int i = n - 2; i >= 0; i--)		{			for(int j = 0; j < triangle[i].size(); j++)			{				dp[j] = triangle[i][j] + std::min(dp[j],dp[j+1]);			}		}		return dp[0];    }};