• [LeetCode] 258. Add Digits 加数字


    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?
      4. You may find this Wikipedia article useful.

    找规律,每9个一循环(1~9),res = (num - 1) % 9 + 1

    数根是将一正整数的各个位数相加(即横向相加),若加完后的值大于10的话,则继续将各位数进行横向相加直到其值小于十为止[1],或是,将一数字重复做数字和,直到其值小于十为止,则所得的值为该数的数根。

    in out   in out
    0   0    10  1
    1   1    11  2
    2   2    12  3
    3   3    13  4
    4   4    14  5
    5   5    15  6
    6   6    16  7
    7   7    17  8
    8   8    18  9
    9   9    19  1

    Java: Trivial and naive

    public class Solution {
        public int addDigits(int num) {
            while (num > 9) {
                num = getInt(num);
            }
            return num;
        }
    
        private int getInt(int num) {
            int result = 0;
            while (num >= 10) {
                result += num % 10;
                num /= 10;
            }
            result += num;
            return result;
        }
    }
    

    Java: T: O(1), S: O(1)

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

    Python: Trivial and naive

    class Solution:
        def addDigits(self, num):
            while num > 9:
                c = 0
                while num:
                    c += num % 10
                    num /= 10
                num = c
            return num

    Python: T: O(1), S: O(1)

    class Solution:
        def addDigits(self, num):
            return (num - 1) % 9 + 1 if num > 0 else 0
    

    C++: Trivial and naive

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

    C++: T: O(1), S: O(1)

    class Solution {
    public:
        int addDigits(int num) {
            return (num - 1) % 9 + 1;
        }
    };
    

    All LeetCode Questions List 题目汇总

      

  • 相关阅读:
    es6 扩展运算符 ...
    回顾2018,展望2019
    vue 兼容IE报错解决方案
    错误的理解引起的bug async await 执行顺序
    js async await 终极异步解决方案
    javascript 之继承-15
    javascript 之原型、原型链-14
    vue入门学习篇——初识vue
    模拟实现select组件功能
    ie8绝对定位存在的坑
  • 原文地址:https://www.cnblogs.com/lightwindy/p/8507117.html
Copyright © 2020-2023  润新知