letecode [258] - Add Digits

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

Example:

Input: 38
Output: 2 
Explanation: The process is like: 3 + 8 = 11, 1 + 1 = 2. 
             Since 2 has only one digit, return it.

Subject to the effect :

　　Given a non-negative num, repeated seeking its digital sum, and to a digital, the digital output.

Understanding:

　　The current digital sum demand num num non-zero when the sum, if the sum is greater than equal to 10, num = sum is updated, the number continues to seek and num. To sum <10.

　　Method two : refer others to practice, to find the law. Num is assumed that three-digit number, the num = 100 * a + 10 * b + c. newNum = a + b + c. Difference dif = num - newNum = 99 * a + 9 * b, is a multiple of 9

　　　　　　I.e. num = newNum + dif. The newNum = num% 9. If newNum = 0, then num = 9.

Code C ++:

class solution {
 public :
     int addDigits ( int num) {
         int sum;
        while (I> = 10 ) { 
            sum = 0 ;
            while (num) { 
                sum + = (whether% 10 ); 
                whether he / = 10 ; 
            } 
            Whether = I; 
        } 
        Return sum; 
    } 
};

operation result:

　　When execution: 12 ms, beat the 74.46% of all users to submit in C ++

　　Memory consumption: 8.3 MB, beat the 5.16% of all users to submit in C ++