• 258. Add Digits


    258. Add Digits

     
     
    Total Accepted: 85971 Total Submissions: 177841 Difficulty: Easy

    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?

    Credits:
    Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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    Code:

    int addDigits(int num) {
        return 1+(num-1)%9;
    }

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  • 原文地址:https://www.cnblogs.com/Alex0111/p/5373628.html
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