LeetCode_258. Add Digits

258. Add Digits

Easy

Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.

Example:

Input: 38
Output: 2 
Explanation: 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?

package leetcode.easy;

public class AddDigits {
	public int addDigits(int num) {
		if (num == 0) {
			return 0;
		} else {
			return ((num - 1) % 9) + 1;
		}
	}

	@org.junit.Test
	public void test() {
		System.out.println(addDigits(38));
	}
}
原文地址:https://www.cnblogs.com/denggelin/p/11765139.html