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?
12 ms
class Solution { public: int addDigits(int num) { return (num - 1) % 9 + 1; } };
8 ms
class Solution { public: int addDigits(int num) { while(true){ if(num<10) return num; else num = reduce(num); } } int reduce(int num){ if(num<10) return num; else return num%10+addDigits(num/10); } };
为什么递归的方法更快呢？
原文：http://www.cnblogs.com/XingyingLiu/p/4977389.html