62.Unique Paths

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).

到达终点要向下走m-1步，向左走n-1步，区别只在于如何排列而已，那么问题很容易转变为排列组合中的问题：现在有x个顺序排列相同的盒子，有y张相同牌，每个盒子只能放一个(x > y)有多少种方式放进去？很显然 C y x =(x−y+１)∗...∗x1∗2...∗y   ,只不过这里的x变为了m - 1 + n - 1,y变为了n - 1。
给出一种实现：

 

class Solution { 

public: 

    int uniquePaths(int m, int n) { 

        if (m > n){ 

            int temp = m; 

            m = n; 

            n = temp; 

        } 

        m = m - 1; 

        n = n - 1; 

        long low = 1; 

        for (int i = 1; i <= m; ++i){ 

            low = i * low; 

        } 

        long high = 1; 

        for (int i = n + 1; i <= m + n; ++i){ 

            high = high * i; 

        } 

        return high/low; 

    } 

};