LeetCode oj 258. Add Digits(数字根）

258. Add Digits

 

Question Editorial Solution

  My Submissions

• Total Accepted: 127957
• Total Submissions: 257201
• Difficulty: Easy

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?

给你一个数，将这个数的各个位相加，直到这个数只有一位

经典的数字根问题，之前做过，就是看能不能除开9，下面给出具体证明

设一个数是abcde，则abcde = a * 10000 + b * 1000 + c * 100 + d * 10 + e,将这个式子拆一下得到(a + b + c + d + e) + (a * 9999 + b * 999 + c * 99 + d * 9),

显然右侧括号的式子可以整除9，所以num % 9的式子成立

需要注意的是特判0 和 9

public int addDigits(int num) {
        if(num % 9 == 0){
            if(num == 0)
              return 0;
            else
              return 9;
        }
        while(num >= 9){
            num %= 9;
        }
        return num;
    }