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.

Example:

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?

Solution

首先要想到這問題就是問%9的結果。 EX: 38 => 11 => 2 == (11%9) == (38%9) 然後問題就單純了，只要濾掉 num!=0 && num%9 == 0 的case(此時應該是9)， 其他就是num%9維結果。

    public int addDigits(int num) {
        if(num == 0) return 0;
        if(num%9 == 0) return 9;
        return num%9;
    }