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.
是一个排列组合的问题,对于一个m*n的grid来说, 解的数目应该是 C m-1/m+n-2
关键就在于怎么算这个解, 如果正常按照公式来算, 那么当m或者n比较大的时候阶乘的数目有可能会大于integer的最大值,于是这里用到组合算法的一个性质:
C k/m = C k-1/m-1 + C k/m-1
在这道题里面的物理意义是, 走到下标为 i,j的方法数目,相当于先走到下表为i-1,j的方格再往下走一步或者是先走到i,j-1的方格然后再往右走一步。所以解题代码如下
public class Solution { public int uniquePaths(int m, int n) { if(m==1||n==1){ return 1; } int [][]r = new int[m][n]; for(int i=0;i<m;i++){ for(int j=0;j<n;j++){ if(i==0||j==0){ r[i][j] = 1; }else{ r[i][j] = r[i-1][j]+r[i][j-1]; } } } return r[m-1][n-1]; } }