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【LeetCode】258. Add Digits 解题小结

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 = 111 + 1 = 2. Since 2 has only one digit, return it.

参考这一篇wiki百科

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

class Solution {public:    int addDigits(int num) {        if (num >= 0 && num < 10) return num;                return (num - 9 * ((num-1)/9));    }};

 

【LeetCode】258. Add Digits 解题小结