Minimum Path Sum

题目描述：

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.
解题思路：
使用动态规划，dp[i][j]表示从（0,0）到（i,j）路径上数字的最小和，由于只能向下和向右移动，

所以状态转移方程为：dp[i][j]=min(dp[i-1][j],dp[i][j-1])+grid[i][j],最后dp[rows-1][cols-1]即为所求。

AC代码如下：

class Solution {
public:
int minPathSum(vector<vector<int>>& grid){
if (grid.size() == 0 || grid[0].size() == 0) return 0;
int m = grid.size();
int n = grid[0].size();
vector<vector<int>> dp(m, vector<int>(n, 0));
//initial the first row
for (int i = 0; i < n; ++i){
if (i == 0) dp[0][i] = grid[0][i];
else dp[0][i] = grid[0][i] + dp[0][i - 1];
}
//initial the first col
for (int i = 0; i < m; ++i){
if (i == 0) dp[i][0] = grid[i][0];
else dp[i][0] = grid[i][0] + dp[i - 1][0];
}

for (int i = 1; i < m; ++i){
for (int j = 1; j < n; ++j){
dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
}
}
return dp[m - 1][n - 1];
}
};