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：
public class Solution {
public int uniquePaths(int m, int n) {
double res = 1;
for (int i = 1; i <= n - 1; i++)
res *= ((double) (m + i - 1) / (double) i);
return (int) Math.round(res);
}
}
提交结果：
