LeetCode刷题笔录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).

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.

这题大概是做LeetCode以来最简单的一道题了。注意的是Robot一开始是在左上角的格子里，而不是在棋盘外面，所以往右和往下能走的步数是m-1和n-1。

于是结果就是在m+n-2个位置里选出m-1个位置用来向右走，剩下的n-1个位置用来向下走。就是(m+n-2)!/(m-1)!(n-1)!.

直接算(m+n-2)!的话可能会溢出，所以直接算(m+n-2)!/(m-1)!，少乘几个数。

public class Solution {
    public int uniquePaths(int m, int n) {
        //the result would be (m-1+n-1)!/(m-1)!(n-1)!
        long numerator = 1;
        int bigger = m > n ? m - 1 : n - 1;
        int smaller = bigger == m - 1 ? n - 1 : m - 1;
        //divide the bigger one's factorial while calculating (m+n)!
        for(int i = m + n - 2; i > bigger; i--){
            numerator *= i;
        }
        long denominator = 1;
        for(int i = 2; i <= smaller; i++){
            denominator *= i;
        }
        return (int)(numerator / denominator);
    }
}