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.

思路：简单的动态规划题目，设f(m, n)为从(0, 0)到达(m, n)的最小和，则 f(m, n) = min(f(m - 1, n), f(m, n - 1)) + A[m][n]

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