LeetCode 258. 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?

把一个数拆成单个，相加，直到加结果为个位数。

解析在不考虑复杂度的情况下，可以使用模拟法循环。

class Solution {public:    int addDigits(int num) {        int result =10;        if(num/10 == 0)            return num;        while(result/10 !=0){            //num = result;            result = 0;             while(num/10 != 0){                result = result + num%10;                num /= 10;            }             result += num;            if(result/10 !=0)            num = result;        }        return result;    }};

在考虑的情况下，找到其规律：(num-1) % 9 + 1，

直接找需要一些数论的知识见：http://my.oschina.net/Tsybius2014/blog/497645

即

public class Solution {    public int addDigits(int num) {       return (num-1) % 9 + 1;    }}

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