标签：
?
| package cn.edu.xidian.sselab.array; /** * * @author zhiyong wang * title: Unique Paths * content: * 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? * */ public class UniquePaths { //第一反应应该是递归, //递归要有约束条件，所以从终点向前进行走，（1,1）到（m,n）的路径和等于（1,1）到（m-1,n）与（m，n-1）路径之和 //总结起来，自己递归的意识还是不够强 //但是这种思路超时了 public int uniquePaths(int m,int n){ if(m==1 || n==1) return 1; else{ return uniquePaths(m-1,n) + uniquePaths(m,n-1); } } //用组合数学的求解方法，从（1，1）到（m,n）的总步数是一定的m + n - 2; //因为只能向右，与向下走，所以向右走的步数也是确定的n-1,向下走的步数也是确定的m-1; //路径总数为：C(m+n-2) (m-1)即（m+n-2）!/(m-1)!(n-1)! public int uniquePath(int m,int n){ int all = n + m - 2; int down = m - 1; double total = 1;//注意这个地方一开始定义是int，会把中间结果处理严重，导致结果不正确 for(int i=1;i<=down;i++){ total = total * (all-down+i) / i; } return (int)total; } } |
标签：
原文地址：http://www.cnblogs.com/wzyxidian/p/5068530.html
