258.Add Digits

tags: `Easy`,`Math`

題目描述

Given an integer num, repeatedly add all its digits until the result has only one digit, and return it.

範例

Example 1:

Input: num = 38
Output: 2
Explanation: The process is
38 --> 3 + 8 --> 11
11 --> 1 + 1 --> 2 
Since 2 has only one digit, return it.

Example 2:

Input: num = 0
Output: 0

Constraints:

0 <= num <= 2³¹ - 1

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

解答

Javascript




function addDigits(num) {
  if (num < 10) return num;
  return addDigits((num % 10) + addDigits(Math.floor(num / 10)));
}

MarsgoatApr 26, 2023



function addDigits(num) {
  return ((num - 1) % 9) + 1;
}

吉神：這題O(1)不是基本嗎？9的倍數檢查方式不是國小教的嗎？？
我道歉嗚嗚嗚，我就爛。
MarsgoatApr 26, 2023

C++






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

Yen-Chi ChenWed, Apr 26, 2023

Python



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

Yen-Chi ChenWed, Apr 26, 2023

C#





public int AddDigits(int num) {
    if (num == 0) return 0;
    int r = num % 9;
    return r == 0 ? 9 : r;
}

JimWed, Apr 26, 2023

$\begin{aligned} x_{n} 10^{n} + x_{n - 1} 10^{n - 1} + . . . + x_{1} 10 + x_{0} \\ = x_{n} (9 + 1)^{n} + x_{n - 1} (9 + 1)^{n - 1} + . . . + x_{1} (9 + 1) + x_{0} \\ = 9 (. . .) + (x_{n} + x_{n - 1} + . . . + x_{1} + x_{0}) \end{aligned}$
忘了怎麼排版…

Java





class Solution {
    public int addDigits(int num) {
        return num != 0 ? (num - 1) % 9 + 1 : 0;
    }
}

Ron ChenMon, Jun 12, 2023

Reference

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