(M)Dynamic Programming:64. Minimum Path Sum

这个题还挺简单的，递归关系很好找，类似 (M)Dynamic Programming:62. Unique Paths。

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