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.

思路如下：

本题是最最基础的动态规划问题，倒三角问题，只能向正下方或者右下方移动，求最短路。

既然是动态规划就秉持着逆推的思想，下一个状态都是由当前状态决定的，所以求出每一

个点对应的最短路径，那么顶点就是所要求的值。

代码如下：

class Solution {

public:

int minimumTotal(vector<vector<int>>& triangle) {

for(int i = triangle.size()-2; i >= 0; i--)

{

for(int j = 0; j < i+1; j++)

{

triangle[i][j] = min(triangle[i+1][j],triangle[i+1][j+1])+triangle[i][j];

}

}

return triangle[0][0];

}

};