Leet Code OJ 258. Add Digits [Difficulty: Easy]
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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 = 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?
代码实现:
public class Solution { public int addDigits(int num) { if(num>9){ return addDigits(num%10+addDigits(num/10)); }else{ return num; } }}
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