题目
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? 
Note: m and n will be at most 100.
分析
这是一道动态规划的题目：
对于一个m∗n矩阵，求解从初始点(0,0)到结束点(m−1,n−1)的路径条数，规定每次只能向右或向下走一步；
我们知道：
1.当i=0,j=[0,n−1]时，f(i,j)=f(i,j−1)=1; 因为每次只能向右走一步，只有一条路径；
2. 当i=[0,m−1],j=0时，f(i,j)=f(i−1,j)=1;因为每次只能向下走一步，只有一条路径；
3. 当(i,j)为其它时，f(i,j)=f(i−1,j)+f(i,j−1)；因为此时可由(i−1,j)向右走一步，或者(i,j−1)向下走一步，为两者之和；
AC代码
//非递归实现回溯，会超时
class Solution {
public:
int uniquePaths(int m, int n) {
if (m == 0 || n == 0)
return 0;
vector<vector<int> > ret(m, vector<int>(n, 1));
//如果矩阵为单行或者单列，则只有一条路径
for (int i = 1; i < m; i++)
for (int j = 1; j < n; j++)
ret[i][j] = ret[i - 1][j] + ret[i][j - 1];
return ret[m-1][n-1];
}
};
递归实现算法（TLE）
class Solution {
public:
int uniquePaths(int m, int n) {
if (m == 0 || n == 0)
return 0;
//如果矩阵为单行或者单列，则只有一条路径
else if (m == 1 || n == 1)
return 1;
else
return uniquePaths(m, n - 1) + uniquePaths(m - 1, n);
}
};