【LeetCode OJ 258】Add Digits

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.

解题思路：思路比较简单，就是求每个位的数之和，判断是否为小于10的数，是就返回结果，不是就继续做求每位数之和的操作。

示例代码：

public class Solution 
{
    public int addDigits(int num)
    {
    	int temp=addSum(num);
    	if(temp<10)
    	{
    		return temp;
    	}
    	else
    		return addDigits(temp);
    }
    /**
     * 求每位的数之和
     * @param num
     * @return
     */
	private int addSum(int num)
	{
		if(num<10)
    		return num;
    	int i=0;
    	int sum=0;
    	while(num!=0)
    	{
    		i=num%10;
    		sum+=i;
    		num=num/10;
    	}
        return sum;
	}
}