1. 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?

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

Note: m and n will be at most 100.

2. 思路

m*n的棋盘，向下是m-1步，向上是n-1步。
在所有的m+n-2步中选择m-1步向下，因此总走法是C(m+n-2, m-1)

本地的复杂度本来应该是组合公式运算的溢出，结果发现标准答案也没考虑溢出问题，试了一下直接计算提交即可AC。

3. 代码

耗时：0ms

class Solution {
public:
    // 一共走（m-1）+（n-1）步, 从其中选择m-1步向下
    // 所以总的走法是C(m+n-2, m-1)
    int uniquePaths(int m, int n) {
        m--; n--;
        long long mn = m + n;
        long long mn_max = max(m, n);
        long long mn_min = min(m, n);
        long long c = 1;
        for (int i = mn_max + 1; i <= mn; i++) {
            c *= i;
        }
        for (int j = 2; j <= mn_min; j++) {
            c /= j;
        }
        return c;
    }
};