258. Add Digits

Description

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?

Solution

Iterative

class Solution {
    public int addDigits(int num) {
        while (num > 9) {
            int digits = 0;
            while (num > 0) {
                digits += num % 10;
                num /= 10;
            }
            num = digits;
        }
        
        return num;
    }
}

Optimized: mod, time O(1), space O(1)

牛掰啊，利用数学规律。注意mod 9为零需要特殊处理。

10^k % 9 = 1
a * 10^k % 9 = a % 9

Then let’s use an example to help explain.

Say a number x = 23456

x = 2* 10000 + 3 * 1000 + 4 * 100 + 5 * 10 + 6
2 * 10000 % 9 = 2 % 9
3 * 1000 % 9 = 3 % 9
4 * 100 % 9 = 4 % 9
5 * 10 % 9 = 5 % 9

Then x % 9 = ( 2+ 3 + 4 + 5 + 6) % 9, note that x = 2* 10000 + 3 * 1000 + 4 * 100 + 5 * 10 + 6

So we have 23456 % 9 = (2 + 3 + 4 + 5 + 6) % 9

class Solution {
    public int addDigits(int num) {
        if (num <= 0) return 0;
        if (num % 9 == 0) return 9;
        return num % 9;
    }
}