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：这道题有两种方法。
第一种是动态规划，构建一个m * n的数组dp, 然后dp[i][j] = dp[i - 1][j] + dp[i][j - 1]。
第二种是公式法，因为一共向下走的操作会有n - 1次，向右走的操作会有m - 1次，因此一共会走m + n - 2次。
然后从m + n - 2次操作中我们要选n - 1次向右走，因此这是一个组合问题。
排列公式 (n, m) = n! / (m!(n - m)!)
实际计算中，我们可以进一步整理为 (n - m + 1) * (n - m + 2) *...* (n - m + m) / m !
这里贴一下动态规划的代码
1 class Solution { 2 public: 3 int uniquePaths(int m, int n) { 4 vector<int> tem(n, 1); 5 vector<vector<int> > dp(m, tem); 6 for (int i = 1; i < m; i++) 7 for (int j = 1; j < n; j++) 8 dp[i][j] = dp[i - 1][j] + dp[i][j - 1]; 9 return dp[m - 1][n - 1]; 10 } 11 };