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：动态规划比较简单的题。用一个二维数组进行存到某个位置的最小值，用两个for i=0~m-1，for j=0~n-1，进行搜索，对两种情况一个j+1，与                    i+1，如果以前的存在大于现在的，特换，如果以前值没有，那就存入。思路比较简单。

class Solution {
public:
    int minPathSum(vector<vector<int> > &grid)  {
        size_t m = grid.size();
        size_t n = grid[0].size();
        int temp[m][n];
        for (size_t i = 0; i != m; i++)
            for(size_t j = 0; j != n; j++)
            temp[i][j] = -1;

        temp[0][0] = grid[0][0];
        for (size_t i = 0; i != m; i++) {
            for(size_t j = 0; j != n; j++) {
              if(j != n-1) {
                if(temp[i][j+1] != -1)
                   temp[i][j+1] =  (temp[i][j+1] > temp[i][j] + grid[i][j+1] ? temp[i][j] + grid[i][j+1] : temp[i][j+1]);
                else
                    temp[i][j+1] =temp[i][j] + grid[i][j+1];
              }
              if(i != m-1) {
                if(temp[i+1][j] != -1)
                   temp[i+1][j] =  (temp[i+1][j] > temp[i][j] + grid[i+1][j] ? temp[i][j] + grid[i+1][j] : temp[i+1][j]);
                else
                    temp[i+1][j] = temp[i][j] + grid[i+1][j];
              }
            }
         }
         return temp[m-1][n-1];
    }
};

LeetCode-Minimum Path Sum