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

Note: m and n will be at most 100.

Example 1:

Input: m = 3, n = 2
Output: 3
Explanation:
From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:

1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right
   Example 2:

Input: m = 7, n = 3
Output: 28

AC代码

class Solution {
public:
    int uniquePaths(int m, int n) {
        int a = m + n - 2, b = min(m, n) - 1, r = b;
        double ans = 1;
        for (int i = 0; i < r; ++i) {
            ans = ans * a / b;
            a -= 1;
            b -= 1;
        }
        return (int)(ans + 0.5);
    }
};

总结

如果用递归，那就是最基本的深搜，可惜会超时，但题目本身是个组合问题，一共会走m+n-2步，向下m-1，向右n-1；计算组合数即可