Given a non-negative integer num, repeatedly add all its digits until the result has only one digit. For example: Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.
题意:给一个非负整数,求其各位数的和只至最后是个位数。
public class Solution { public int addDigits(int num) { ////循环和递归解决~ int sum=0; if(num/10==0)return num; while(num!=0){ sum=sum+num%10; num=num/10; } // System.out.println(sum); return addDigits(sum); } }
PS:额,直接循环加递归解决。。。。
进阶:不使用循环和递归。
Leetcode 258. Add Digits JAVA语言
原文:http://fulin0532.blog.51cto.com/6233825/1890563