258. Add Digits

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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?

num = a * 10000 + b * 1000 + c * 100 + d * 10 + e

即：num = (a + b + c + d + e) + (a * 9999 + b * 999 + c * 99 + d * 9)

1 public class Solution {
2     public int addDigits(int num) {
3         return (num-1)%9 + 1;
4     }
5 }

258. Add Digits

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原文地址：http://www.cnblogs.com/guoguolan/p/5383492.html