leetcode: Add Digits

标签：

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.

Follow up:
Could you do it without any loop/recursion in O(1) runtime?

题目描述：给出一个数组，不断地分割数字，直到这个数字是一个数字为止。题目很简单，直接上代码。

代码实现：

class Solution {
public:
    int addDigits(int num) {
        while(num<10)
        {
            return num;
        }
        int s=0;
        while(num>0)
        {
            s+=num%10;
            num/=10;
        }
        return addDigits(s);
    }
};

leetcode: Add Digits

标签：

原文地址：http://blog.csdn.net/flyljg/article/details/47835045