动态规划（dynamic programming）之：minimum-path-sum

从数组的左上角到右下角，求经过的路径的最小和。

思路：

（1）状态设定：f[x][y]为从坐标（0,0）走到（x,y）的最短路径和

（2）状态方程为：f[x][y] = (x,y) + min{f[x-1][y], f[x][y-1]};

（3）初始化：f[0][0] = grid[0][0]；f[i][0] = sum(0,0 —> i,0)；f[0][i] = sum(0,0 —> 0,i)

（4）结果：f[m-1][n-1]; 

class Solution {
public:
    int minPathSum(vector<vector<int> > &grid) {
        if(grid.empty() || grid[0].empty())
            return 0;
        int m = grid.size();
        int n = grid[0].size();
        vector<vector<int> > ret(m, vector<int>(n,0));
        ret[0][0] = grid[0][0];
        
        for(int i = 1; i < m; ++i)
            ret[i][0] = grid[i][0] + ret[i-1][0];
        
        for(int i = 1; i < n; ++i)
            ret[0][i] = grid[0][i] + ret[0][i-1];
        
        for(int i = 1; i < m; ++i){
            for(int j = 1; j < n; ++j){
                ret[i][j] = grid[i][j] + min(ret[i-1][j], ret[i][j-1]);
            }
        }
        return ret[m-1][n-1];
    }
};

上面是没有对grid进行修改，下面这个是直接在grid上面做手脚

int minPathSum(vector<vector<int> > &grid) {
        if(grid.empty() || grid[0].empty())
            return 0;;
        int M = grid.size();
        int N = grid[0].size();
        for(int i = 0; i < M; ++i){
            for(int j = 0; j < N; ++j){
                if(i == 0 && j != 0)    grid[i][j] += grid[i][j-1];
                if(i != 0 && j == 0)    grid[i][j] += grid[i-1][j];
                if(i*j != 0)
                    grid[i][j] += min(grid[i-1][j], grid[i][j-1]);
            }
        }
        return grid[M-1][N-1];
    }