LeetCode 62 _ Unique Paths 全部不同路径

Description: 

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

Solution: 

这道题让我们求出从规格为m*n的方格左上角走到右下角的所有不同路径总数

这道题很简单，我们知道，如果不走回头路，那么机器人只能向右或向下走，7*3的方格实际上要向右走6次，向下走2次

所以该问题本质是“m个A，n个B，有多少种不同排列”的高中概率论问题

求法即为在（m+n）个空位中选择m个空位给A，剩下的即为B的位置

答案即C62，通用解为(m+n)!/(m!*n!)

由于中间可能在乘分子的时候超过范围，因此中间值sum最好取double，以使结果的范围更大

Code:

public int uniquePaths(int m, int n) {
    if (m <= 0 && n <= 0){
        return 0;
    }

    int total = m + n - 2;
    int less = Math.min(m,n) - 1;
    double sum = 1;
    for (int i = 1; i <= less; i++){
        sum = sum * (total + 1 - i) / i;
    }
    return (int)sum;
}

提交情况：

Runtime: 0 ms, faster than 100.00% of Java online submissions for Unique Paths.

Memory Usage: 31.8 MB, less than 100.00% of Java online submissions for Unique Paths.

原文地址：https://www.cnblogs.com/zingg7/p/10655801.html