Minimum Path Sum

原题：
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：
这个明显是一个DP的问题，从（0,0）点出发，到达（m-1, n-1）点，每次选择的时候都是依据上一次的结果推断后做累加。
dp[i][j] = (dp[i-1][j] > dp[i][j-1] ?

dp[i][j-1] : dp[i-1][j]) + grid[i][j];
选择最小的值来更新当前的结果。

能够AC的C++代码例如以下：

   int minPathSum(vector<vector<int>>& grid) {
        if(grid.size() < 1 || grid[0].size() < 1)
            return -1;

        int m = grid.size(), n = grid[0].size();
        int **dp = new int*[m];
        for(int i=0; i<m; i++){
            dp[i] = new int[n];
            memset(dp[i], 0, sizeof(int) * n);
        }

        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] = (dp[i-1][j] > dp[i][j-1] ? dp[i][j-1] : dp[i-1][j])  + grid[i][j];
            }
        }

        return dp[m-1][n-1];
    }