[258]Add Digits

【题目描述】

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.

【思路】

无

【代码】

class Solution {public:    int addDigits(int num) {        int sum=0;        if(num<=9) return num;        sum=getsum(num);        while(sum>=10){            sum=getsum(sum);        }        return sum;    }    int getsum(int n){        int ans=0;        while(n){            ans+=n%10;            n=n/10;        }        return ans;    }};

在discuss里看到一种纯数学的方法，可惜并米有效率高点还是要耗时8ms。。不过贴上去借鉴一下。。

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