Leetcode_258_AddDigits

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Add Digits

Discription：

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?

Solution:

1 class Solution {
2 public:
3     int addDigits(int num) {
4         return 1+(num-1)%9;
5     }
6 };

Hints:

　digital root: https://en.wikipedia.org/wiki/Digital_root

　vedic square: https://en.wikipedia.org/wiki/Vedic_square

　use the formular as follows:

Leetcode_258_AddDigits

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原文地址：http://www.cnblogs.com/nmj93/p/4900542.html