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?

分析：这道题可以用动态规划解，f[i][j]表示从(0,0)到(i,j)的路径数，递推公式为f[i][j] = f[i-1][j] + f[i][j-1]。时间复杂度O(n^2)，空间复杂度O(n^2)。代码如下：

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

 其实在用动态规划时，我们可以用一个滚动数组代替二维数组。代码如下：

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