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题目链接:https://leetcode.com/problems/add-digits/
题目:
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:给定一个非负整数num,将其个位重复相加直到其结果仅有一位为止
题目很简单,直接贴代码吧~
public class Solution { public int addDigits(int num) { int result = 0; int gewei = 0; if(num/10 == 0) return num; while(num != 0) { gewei = num % 10; result += gewei; num /= 10; } if(result >= 10) return addDigits(result); else return result; } }
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原文地址:http://blog.csdn.net/yangyao_iphone/article/details/47984707