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.
思路:
(a + b + c + d) % 9
= (999a + 99b + 9*c + a + b + c + d) % 9
= abcd % 9
public int addDigits(int num){
if(num ==0)return0;
return num%9==0?9:num%9;
}
原创声明:本文系作者授权腾讯云开发者社区发表,未经许可,不得转载。
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原创声明:本文系作者授权腾讯云开发者社区发表,未经许可,不得转载。
如有侵权,请联系 cloudcommunity@tencent.com 删除。