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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.
代码:
public class Solution { public int addDigits(int num) { return 1+(num-1)%9; } }
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原文地址：http://www.cnblogs.com/finalboss1987/p/5079065.html
