258. Add Digits-优快云博客

本文介绍了一种高效的数字压缩算法，该算法能在O(1)时间内计算出任意非负整数不断累加其各位数字直至结果仅剩一位数的过程。通过数学推导发现，此过程等价于对原数减一后取九的模数再加一。

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:

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.
```
class Solution {
public:
    int addDigits(int num) {
       return 1+(num-1)%9;
    }
};
```
思路是：假设一个数是abcde，则abcde=a*10000+b*1000+c*100+d*10+e=(a+b+c+d+e)+(a*9999+b*999+c*99+d*9)，已知后面括号里的数一定是能够被9整除的，所以abcde和a+b+c+d+e模9所得的余数相同。因为a+b+c+d+e结果依然可能不是只有一位数，所以循环下去，可是这个规律依然存在。但有个例外就是当数为9时，取模后为0，但其实是9的，只要将数先减小1，最后结果再增加1即可