Add Digits

• Add Digits
  • 题目描述
  • 解法

Add Digits

题目描述

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?

解法

代码如下:

public static int addDigits( int num ) {
    return 1 + ( num - 1 ) % 9;
}

思考过程: 具体介绍请看Digital root–wiki

时空复杂度: 时间复杂度是O(1), 空间复杂度是O(1)