[LeetCode] 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.

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;}
};

转载于:https://www.cnblogs.com/vincently/p/4803875.html

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