258. Add Digits

问题

Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.

Could you do it without any loop/recursion in O(1) runtime?

例子

Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.

分析

1=>1 2=>2 3=>3 4=>4 5=>5 6=>6 7=>7 8=>8 9=>9 10=>1 12=>2...
似乎可以看出，结果就是余9的数字。用35478验证一下，结果是9，余9是0。不对，应该要对9的倍数特殊对待：9的倍数的结果就是9.

要点

找规律

时间复杂度

O(1)

空间复杂度

O(1)

代码

分类讨论

class Solution {
public:
    int addDigits(int num) {
        return num != 0 && num % 9 == 0 ? 9 : num % 9;
    }
};

合并情况

class Solution {
public:
    int addDigits(int num) {
        return (num - 1) % 9 + 1;
    }
};