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[LeetCode] #66 Plus One (Easy)

LeetCode 第 66 題 Plus One,難度 Easy

Ray

LeetCode #66 run result

用 Python3 解 LeetCode 系列,Plus One,屬於 Easy

原始題目

You are given a large integer represented as an integer array digits, where each digits[i] is the ith digit of the integer. The digits are ordered from most significant to least significant in left-to-right order. The large integer does not contain any leading 0’s.

Increment the large integer by one and return the resulting array of digits.

Example 1:

Input: digits = [1,2,3]
Output: [1,2,4]
Explanation: The array represents the integer 123.
Incrementing by one gives 123 + 1 = 124.
Thus, the result should be [1,2,4].

Example 2:

Input: digits = [4,3,2,1]
Output: [4,3,2,2]
Explanation: The array represents the integer 4321.
Incrementing by one gives 4321 + 1 = 4322.
Thus, the result should be [4,3,2,2].

Example 3:

Input: digits = [0]
Output: [1]
Explanation: The array represents the integer 0.
Incrementing by one gives 0 + 1 = 1.
Thus, the result should be [1].

Example 4:

Input: digits = [9]
Output: [1,0]
Explanation: The array represents the integer 9.
Incrementing by one gives 9 + 1 = 10.
Thus, the result should be [1,0].

題目分析

  • 傳入一個純正整數的陣列
  • 該陣列是一個大正整數,依照每個位數拆開
  • 計算出該正整數 +1 的結果
  • 把結果依照輸入的方式也輸出成陣列

解題過程

  1. 把輸入的陣列數字一個一個串成字串
  2. 字串轉換成數字,然後 +1
  3. 每個位元拆開做成新的 list 後回傳
class Solution:
    def plusOne(self, digits: List[int]) -> List[int]:
        number_string = ''
        number_new = []

        # list 用字串串起來
        for number in digits:
            number_string += str(number)

        # 字串轉數字,數字 +1
        number_string = str(int(number_string) + 1)

        # 每一個位元拆開成新的 list
        for number in number_string:
            number_new.append(number)
            
        return number_new

結果

result

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