258. Add Digits规律

Posted 2020-09-06

2017/3/16 22:36:02

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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?

 

版本1    循环

1. publicclassSolution{
2. publicint addDigits(int num){
3. while( num /10!=0)
4. num = num/10+ num%10;
5. return num;
6. }
7. }

 

版本2   列出表格观察，发现9个数字一循环的规律，一行代码就搞定。

1. publicclassSolution{
2. publicint addDigits(int num){
3. return num %9==0? num ==0?0:9: num%9;
4. }
5. }