Leetcode 258. Add Digits (number theory)

Given a non-negative integer num , repeatedly add all its digits until the result have only one digit.

For example:

Given num = 38 , the process is like: 3 + 8 = 11 , 1 + 1 = 2 . Since have only one 2 digit, return it.

The problem is to find the number root of a number.

The number root has a congruence property: A number with its number root pair (b-1) congruence (b is the binary number).

A simple example of this is clear:

123=1*100+2*10+3

=1* (99+1) +2* (9+1) +3

= (99+2*9) + (1+2+3)

The previous item can be divisible by 9, and the next one is the number of each bit. The number that is obtained after the 1+2+3, still can be so split, go down until the number root.

Therefore a number with its number root pairs (b-1) congruence (b is the binary number).

For the subject, we use this property for the number of roots, because the number of roots is a number, and the number of%9 and num%9 results the same, so we are directly num%9, but the number we find here is not number root, the number is [0,9], and%9 to find out is [1,8], So we add a small processing technique: minus 1 First, and then add 1 after the mold.

class Solution {public:    int adddigits (int  num) {          return1+ (num-1)%9;       }};

Leetcode 258. Add Digits (number theory)