64. Minimum Path Sum

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.Note: You can only move either down or right at any point in time.Example 1:
[[1,3,1],
[1,5,1],
[4,2,1]]
Given the above grid map, return 

7

. Because the path 1→3→1→1→1 minimizes the sum.
好久没做动态规划了，快忘光了。
注意边缘与中心是不同的，写出状态转移方程就好了。
代码：class Solution {
public:
int minPathSum(vector<vector<int>>& grid) {
int m=grid.size(),n=grid[0].size();
for(int i=1;i<m;i++)
grid[i][0]+=grid[i-1][0];
for(int j=1;j<n;j++)
grid[0][j]+=grid[0][j-1];
for(int i=1;i<m;i++)
{
for(int j=1;j<n;j++)
{
grid[i][j]=min(grid[i-1][j],grid[i][j-1])+grid[i][j];
}
}
return grid[m-1][n-1];
}
};