leetcode/258. Add Digits

//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?

solution 1

int addDigits(int num) {    if(num) {        int res = num % 9;        return (res == 0)? 9:res;    } else {        return num;    }}

solution 2

int addDigits(int num) {    int res = num % 9;    return (res != 0 || num == 0)? res:9;}

solution 3

int addDigits(int num) {        return (num - 1) % 9 + 1;}

//根据同余定理10^n ≡ 1 (mod 9)得a_n10^n + … + a_110 + a_0 ≡ a_n + … + a_1 + a_0 (mod 9)
//其中a_n表示位数的系数，综上所述，取模后能够得到各个位数系数和，且这系数和是小于9的整数
//当且仅当n对9取模为零时侯要分开考虑，n%9 = 0 ,当给的num是0是，返回值才为0，否则返回9