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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.

********************************************/ 参考别的程序，思想为：

he problem, widely known as digit root problem, has a congruence formula:

For base b (decimal case b = 10), the digit root of an integer is:

dr(n) = 0 if n == 0

dr(n) = (b-1) if n != 0 and n % (b-1) == 0 dr(n) = n mod (b-1) if n % (b-1) != 0 or

dr(n) = 1 + (n - 1) % 9

Note here, when n = 0, since (n - 1) % 9 = -1, the return value is zero (correct).

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

自己的程序，容易理解：

int addDigits(int num) {       while (num>9)            num=num/10+num%10;        return num;    }

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