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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?
Hint:
Tags: Math
Similar Problems: (E) Happy Number
不用 循环 与 递归,毫无思路,从最简单情况开始分析:
1 class Solution { 2 public: 3 int addDigits(int num) 4 { 5 return 1 + (num-1)%9; 6 } 7 }
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原文地址:http://www.cnblogs.com/whl2012/p/5596675.html