[leetcode][math] Add Digits

本文探讨了一个有趣的问题:如何将一个非负整数的所有位数反复相加直至只剩一位数,并提出了两种解决方案。一种是通过循环迭代的方式实现,另一种则是在O(1)时间内通过数学方法直接得出结果。

题目:

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 = 111 + 1 = 2. Since 2 has only one digit, return it.

Follow up:
Could you do it without any loop/recursion in O(1) runtime?

Hint:

  1. A naive implementation of the above process is trivial. Could you come up with other methods?

  2. What are all the possible results?
  3. How do they occur, periodically or randomly?
class Solution {
public:
    int addDigits(int num) {
        int res = 0;
        bool doneFlag = false;
        while(1){
            res = 0;
            while(num > 0){
                res += num%10;
                num /= 10;
            }
            if(res/10 == 0) break;
            num = res;
        }
        return res;
    }
};

class Solution {
public:
    int addDigits(int num) {
        if(num == 0) return 0;
        if(num % 9 == 0) return 9;
        return num%9;
    }
};


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值