GuilinDev

Lc0258

05 August 2008

258 Add Digits

原题概述

Given a non-negative integer 

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num
, repeatedly add all its digits until the result has only one digit.

Example:

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Input: 38
Output: 2 
Explanation: 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?

题意和分析

这个题比较简单，follow up是常数级的复杂度；如果用正常的loop或者递归的话是线性时间复杂度，因此需要略微取巧。

代码

正常的loop

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class Solution {
    public int addDigits(int num) {
        if (num < 10) {
            return num;
        }
        int temp = num;
        int result = 0;
        while (true) {
            while (temp != 0) {
                result += temp % 10;
                temp /= 10;
            }
            if (result < 10) {
                return result;
            } else {
                temp = result;
                result = 0;
            }
        }
    }
}

递归

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class Solution {
    public int addDigits(int num) {
        if (num < 10) {
            return num;
        }
        return addDigits(num % 10 + num / 10);
    }
}

接下来看常数的解法，对9进行取模,当作9进制来算（证明略），参考这里

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class Solution {
    public int addDigits(int num) {
        if (num < 10) {
            return num;
        } else if (num % 9 == 0) {
            return 9;
        } else {
            return num % 9;
        }
    }
}
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