LeetCode-62-Unique Paths

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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?

题意：就是一个m*n的棋盘的上，从一个位置到另一位置的最短路径的个数。每次只能向下或向右走一步。

思路：

　　其实就是个高中的组合数学的问题。

　　m*n的棋盘，一共需要走(m-1)+(n-1)步，向右走m-1步，向下走n-1步，这(m-1)+(n-1)步中，只要确定了哪些步向右，即同时确定了哪些步向下走，反之亦然。

　　答案即C(m+n-2,m-1)或C(m+n-2,n-1)

Java代码如下：

1 public class Solution {
2     public int uniquePaths(int m, int n) {
3         double res = 1;
4         for (int i = 1; i <= n - 1; i++) 
5             res *= ((double) (m + i - 1) / (double) i);
6         return (int) Math.round(res);
7     }
8 }

提交结果：

LeetCode-62-Unique Paths

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原文地址：http://www.cnblogs.com/z941030/p/5609042.html