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?

不用循环

 xyz = 100x + 10y +z, xyz % 9 = 99*x + 9*y + (x + y + z) % 9 = (x + y + z) % 9

java(3ms):

 1 public class Solution {
 2     public int addDigits(int num) {
 3         if (num == 0)
 4             return 0 ;
 5         if (num > 9)
 6             num %= 9 ;
 7         if (num == 0)
 8             num = 9 ; 
 9         return num;
10     }
11 }

258. Add Digits

标签：strong   dig   java   its   div   blog   ddd   turn   one   

原文地址：http://www.cnblogs.com/-Buff-/p/6024822.html