[LeetCode]64. Add Digits数根

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?

Hint:

1. A naive implementation of the above process is trivial. Could you come up with other methods?
2. What are all the possible results?
3. How do they occur, periodically or randomly?
4. You may find this Wikipedia article useful.

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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解法1：利用to_string库函数，只要输入整数的数字个数大于1，则执行自身数字相加的过程。

class Solution {
public:
    int addDigits(int num) {
        string s = to_string(num);
        while (s.size() > 1) {
            int tmp = 0;
            for (int i = 0; i < s.size(); ++i)
                tmp += s[i] - '0';
            s = to_string(tmp);
        }
        return stoi(s);
    }
};

或者直接分别取出某位累加：

class Solution {
public:
    int addDigits(int num) {
        while (num >= 10) {
            int sum = 0;
            while (num > 0) {
                sum += num % 10;
                num /= 10;
            }
            num = sum;
        }
        return num;
    }
};

解法2：参考维基百科Digital root，

It helps to see the digital root of a positive integer as the position it holds with respect to the largest multiple of 9 less than it. For example, the digital root of 11 is 2, which means that 11 is the second number after 9. Likewise, the digital root of 2035 is 1, which means that 2035 − 1 is a multiple of 9. If a number produces a digital root of exactly 9, then the number is a multiple of 9.

With this in mind the digital root of a positive integer  may be defined by using floor function , as

不难写出代码：

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