problem：
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.
thinking：
考察DP，开一个n(三角型矩阵的最后一行的大小)大小的数组，每次记录上一行到下一行以该元素为终点的路径和，
这样，最后在数组中寻找最小的一个元素就是最小路径和
code：
class Solution {
public:
int minimumTotal(vector<vector<int> > &triangle) {
if (triangle.size() == 0)
return 0;
vector<int> f(triangle[triangle.size()-1].size());
f[0] = triangle[0][0];
for(int i = 1; i < triangle.size(); i++)
for(int j = triangle[i].size() - 1; j >= 0; j--)
if (j == 0)
f[j] = f[j] + triangle[i][j];
else if (j == triangle[i].size() - 1)
f[j] = f[j-1] + triangle[i][j];
else
f[j] = min(f[j-1], f[j]) + triangle[i][j];
int ret = INT_MAX;
for(int i = 0; i < f.size(); i++)
ret = min(ret, f[i]);
return ret;
}
};原文地址：http://blog.csdn.net/hustyangju/article/details/45305935
