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.
分析一:最简单的循环方法
class Solution { public:int addDigits(int num) {while (num / 10 != 0) {int temp = 0;while (num != 0) {temp += num % 10;num /= 10;}num = temp;}return num;} };
分析二:参考https://leetcode.com/discuss/52122/accepted-time-space-line-solution-with-detail-explanations
class Solution { public:int addDigits(int num) {return 1 + (num - 1) % 9;} };