Unique Paths

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?

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

Note: m and n will be at most 100.

此题思路有3种！

1、直接数学公式，用组合的方法，直接用(M+N-2)!/(M-1)!*(N-1)!  意思是 水平方向有 n-1 中选择，垂直方向 m-1中选择。所以总体有 m+n-2，只需在这m+n-2种方式选择向右走 n-1 步或者 向下的m-1  步即可。注意实现 (n-m)!的函数即可，注意溢出问题！

2、利用DFS，每个地方都有俩种可能，依次即可，记得带备忘录（缓存数据）-提高速度。

3、可知此题涉及子问题重叠，故而会想到DP算法，以下是实现。

dp[i][j] 表示从[0][0]到[i][j] 的所有路径之和。有以下转移方程：dp[i][j] = dp[i-1][j] + dp[i][j-1];

处理上述时，注意到 i = 0，j = 0需特殊处理！

   int uniquePaths(int m, int n)     {        return count_paths(m,n);    }        int count_paths(int m,int n)   {vector<vector<int>> dp(m,vector<int>(n,0));//dp[][]dp[0][0] = 1;for (int i = 1;i < n;i++){dp[0][i] = 1;}for (int i = 1;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];}