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，这样就可以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]; } };