258. Add Digits

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Problem:

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?

A possible solution:

int addDigits(int num) {
    int temp;
    while(num >= 10)
    {
        while(num >= 10)
        {
            temp += num%10;
            num = num/10;
        }
        num  += temp; 
        temp = 0;
    }
    return num;
}

　more efficient way:

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

258. Add Digits

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原文地址：http://www.cnblogs.com/liutianyi10/p/5517753.html