4/26每日一題

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

 

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?

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class Solution(object):

    def addDigits(self, num):

        """

        :type num: int

        :rtype: int

        """

        '''判斷是否是1位數,如果是就返回num'''

        if len(str(num)) == 1:

            return num

        else:

           '''如果非1位數,將各位數加總,返回到num再次判斷是否為1位數'''

           while len(str(num)) >1 :

                str_num=str(num)

                lis_num=list(str_num)

                tol = sum(map(int,lis_num))

                #map(func,iterable)

                '''func是用來對可迭代的函數進行處理

                這邊將list中每個元素轉成int後進行加總

                '''

                num = tol

                #重新賦予num新的值,再次進入while判斷是否為一位數

               

        return num

               

           

           

s=Solution()

print(s.addDigits(38))

print(s.addDigits(0))

print(s.addDigits(1))

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PS C:\Users\K\OneDrive\桌面\AUTOKI> & C:/Users/K/AppData/Local/Programs/Python/Python311/python.exe c:/Users/K/OneDrive/桌面/AUTOKI/think.py 2 0 1 PS C:\Users\K\OneDrive\桌面\AUTOKI>

"""最佳解"""

"""

Time Complexity :- BigO(1)

Space Complexity :- BigO(1)

"""

class Solution:

    def addDigits(self, num: int) -> int:

        if num == 0 : return 0

        if num % 9 == 0 : return 9

        else : return (num % 9)

"""

複雜

時間複雜度：O(logn^2)

用於遞歸迭代數字

空間複雜度：O(1)

"""

class Solution:

    def addDigits(self, num: int) -> int:

        while num > 9:

            sum = 0

            while num:

                sum += num%10

                num = num//10

               

            num = sum

        return num