A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

How many possible unique paths are there?

思路: 如果使用递归，那么DFS(i,j)=v[i][j]+max{DFS(i,j-1), DFS(i-1,j)}，问题是DFS(i,j)可能会被计算两次，一次来自它右边的节点，一次来自它下面的节点。每个节点都被计算两次，时间复杂度就是指数级的了。解决方法是使用动态规划存储节点信息，避免重复计算。

Above is a 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

class Solution {public:    int uniquePaths(int m, int n) {        int dp[m][n];                dp[0][0] = 1;        for(int i = 0; i< n; i++ )        {            dp[0][i] = 1;        }        for(int i = 0; i< m; i++ )        {            dp[i][0] = 1;        }                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];            }        }        return dp[m-1][n-1];    }};