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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.
分析一：最简单的循环方法
class Solution { public: int addDigits(int num) { while (num / 10 != 0) { int temp = 0; while (num != 0) { temp += num % 10; num /= 10; } num = temp; } return num; } };
分析二：参考https://leetcode.com/discuss/52122/accepted-time-space-line-solution-with-detail-explanations
class Solution { public: int addDigits(int num) { return 1 + (num - 1) % 9; } };
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原文地址：http://www.cnblogs.com/vincently/p/4803875.html
