美文网首页
258. Add Digits

258. Add Digits

作者: YellowLayne | 来源:发表于2017-10-30 14:45 被阅读0次

1.描述

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?

2.分析

3.代码

class Solution {
public:
    int addDigits(int num) {
        if (num / 10 == 0) return num;
        int newnum = 0;
        while (num / 10) {
            newnum += num % 10;
            num /= 10;
        }
        newnum += num;
        return addDigits(newnum);
    }
};

相关文章

  • Leetcode PHP题解--D69 258. Add Dig

    D69 258. Add Digits 题目链接 258. Add Digits 题目分析 给定一个数字,给每一位...

  • 2019-02-02

    LeetCode 258. Add Digits Description Given a non-negative...

  • 258. Add Digits

    258. Add Digits[思路]数字累加,将给定的一个整数,将个位,十位,百位等相加,连续操作,直到最后的值...

  • 258. Add Digits

    循环: 这个办法很神奇

  • 258. Add Digits

    Problem Given a non-negative integer num, repeatedly add ...

  • 258. Add Digits

    问题 Given a non-negative integer num, repeatedly add all i...

  • 258. Add Digits

    1.描述 Given a non-negative integer num, repeatedly add all...

  • 258. Add Digits

    C++ Java Javascript 题目问能不能做出不用循环和递归的O(1)复杂度,没做出来,(╯‵□′)╯︵...

  • 258. Add Digits

    Given a non-negative integernum, repeatedly add all its d...

  • 258. Add Digits

    传统的loop方法 我有点讨厌的trick方法: 对于B进制来说 N最后遗留下来的数符合以下规则:1.如果N=0,...

网友评论

      本文标题:258. Add Digits

      本文链接:https://www.haomeiwen.com/subject/ziwppxtx.html