LeetCode(43) Add Digits

题目描述

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.

题目要求对一个非负数的各个数位上的数字求和 ，直到最终和为一个个位数为止。

解题思路与代码

本题最简单的方法就是根据题目的意思进行循环操作，直到最终得到的数字只有一位为止。

class Solution {
public:

    int sumOfDigits(int num)
    {
        int sum = 0;
        while(num)
        {
            sum += num % 10;
            num = num / 10;
        }
        return sum;
    }
    int addDigits(int num) {
        while(num / 10 != 0)
        {
            num = sumOfDigits(num);
        }
        return num;
    }
};

解法进阶

在LeetCode上我们可以看到本题的提示：
Could you do it without any loop/recursion in O(1) runtime?

Hint:

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

这里wiki上对本问题的数学解法进行了讨论，我们不按照这个也可以使用之前的代码得到的计算结果去寻找一些规律。这里我打印出了1~50时的结果，可以看出得到的结果是1~9的循环。
num: 1, result: 1
num: 2, result: 2
num: 3, result: 3
num: 4, result: 4
num: 5, result: 5
num: 6, result: 6
num: 7, result: 7
num: 8, result: 8
num: 9, result: 9
num: 10, result: 1
num: 11, result: 2
num: 12, result: 3
num: 13, result: 4
num: 14, result: 5
num: 15, result: 6
num: 16, result: 7
num: 17, result: 8
num: 18, result: 9
num: 19, result: 1
num: 20, result: 2
num: 21, result: 3
num: 22, result: 4
num: 23, result: 5
num: 24, result: 6
num: 25, result: 7
num: 26, result: 8
num: 27, result: 9
num: 28, result: 1
num: 29, result: 2
num: 30, result: 3
num: 31, result: 4
num: 32, result: 5
num: 33, result: 6
num: 34, result: 7
num: 35, result: 8
num: 36, result: 9
num: 37, result: 1
num: 38, result: 2
num: 39, result: 3
num: 40, result: 4
num: 41, result: 5
num: 42, result: 6
num: 43, result: 7
num: 44, result: 8
num: 45, result: 9
num: 46, result: 1
num: 47, result: 2
num: 48, result: 3
num: 49, result: 4
num: 50, result: 5

依据上述规律我们可以写出简单的代码：

class Solution {
public:
    int addDigits(int num) {
        if(!num) return 0;
        int a = num % 9;
        if (a)  return a;
        else    return 9;
    }
};