leetcode - Add Digits

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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?

class Solution {
public:
    int addDigits(int num) { 
        if(num==0) return 0;
        if(num%9 == 0) return 9;
        return num%9;
    }
};

找规律：https://en.wikipedia.org/wiki/Digital_root

leetcode - Add Digits

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原文地址：http://www.cnblogs.com/shnj/p/4739342.html