algorithms/cpp/addDigits/addDigits.cpp

// Source : https://leetcode.com/problems/add-digits/
// Author : Timothy Lim, Hao Chen
// Date   : 2015-10-1

/********************************************************************************** 
 * 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?
 *
 **********************************************************************************/

class Solution {
    public:

        int addDigits(int num) {
            switch(random()%5+1){
                case 1: return addDigits01(num);
                case 2: return addDigits02(num);
                case 3: return addDigits03(num);
                case 4: return addDigits04(num);
                default: return addDigits05(num);
            }

        }

        //regualr way
        int addDigits01(int num) {
            while(num > 9) {
                int sum;
                for(sum=0; num > 0; sum += num%10 , num/=10);
                num = sum;
            }
            return num;

        }

        //This solution looks is very tricky, but acutally it is easy to understand.
        //it just keep adding the last digital until the num < 10
        int addDigits02(int num) {
            while(num > 9) {
                num = num / 10 + num % 10;
            }
            return num;

        }

        // Let's observe the pattern
        //    1    1
        //    2    2
        //    ... ...
        //    8    8    
        //    9    9    
        //    10    1
        //    11    2
        //    12    3    
        //    ... ...
        //    17    8
        //    18    9
        //    19    1
        //    20    2
        //    ...  ...
        // It looks most of number just simply %9 is the answer, 
        // but there are some edge cases.
        //    9%9=0 but we need 9. 
        //    18%9=0 but we need 9
        // so we can find the solution is:
        //    1) num <=9, return num
        //    2) num > 9, reutrn num%9 if num%9>0
        //                return 9 if num%9 ==0
        int addDigits03(int num) {
            return num >9 ? ((num %9)==0 ? 9:num%9) : num;
        }

        //But actually, we can use (num-1)%9 + 1 to make all cases right.
        int addDigits04(int num){
            return (num - 1) % 9 + 1;
        }

        //This solution is similar with pervious solution.
        int addDigits05(int num){
            return num - 9 * ((num - 1)/9);
        }

};