Epoch

題目

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?

想法

把數字拆解，加起來，超過10的話就要重複步驟，可用迴圈或遞迴，O(1)解要另外想。

code

while loop

int addDigits(int num) {
    int digit = 0;
    while (1) {
        if (num > 0) {
            digit += num%10;
            num/=10;
        }
        else if (num == 0 && digit >= 10){
            num = digit;
            digit = 0;
        } else if (digit < 10) {
            break;
        }
    }
    return digit;
}

recursive

int re(int num) {
    if (num <10) {
        return num;
    } else {
        return re(num%10+num/10);
    }
}
int addDigits(int num) {
    return re(num);
}

O(1)

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