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.

思路：这题首先想到的是递归，可是递归效率太慢。超时不解释。

于是换动态规划，也算典型的动态规划的题了。

详细代码例如以下：

public class Solution {
    public int uniquePaths(int m, int n) {
        
        //本题解法为动态规划
        //状态转移方程f[i][j] = f[i-1][j] + f[i][j-1];
        //f[i][j]的值即为路径的数量
        
        int[][] f = new int[m][n];
        for(int i = 0; i < m ; i++)//第一列赋值为1
            f[i][0] = 1;
        for(int i = 0; i < n; i++)//第一行赋值为1
            f[0][i] = 1;
            
        for(int i = 1; i < m ; i++)
            for(int j = 1; j < n ; j++){
            	 f[i][j] = f[i-1][j] + f[i][j-1];
            	 //System.out.println("f[" + i +"][" + j+ "]" + f[i][j]);
            }
        return f[m-1][n-1];//返回结果值
    }
}