[24] 用 python 刷 Leetcode: 66 plus-one

原始题目

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].

题目分析

传入一个纯正整数的阵列
该阵列是一个大正整数，依照每个位数拆开
计算出该正整数 +1 的结果
把结果依照输入的方式也输出成阵列

解题过程

把输入的阵列数字一个一个串成字串
字串转换成数字，然後 +1
每个位元拆开做成新的 list 後回传

class Solution:
    def plusOne(self, digits: List[int]) -> List[int]:
        number_string = ''
        number_new = []

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

        # 字串转数字，数字 +1
        number_string = str(int(number_string) + 1)

        # 每一个位元拆开成新的 list
        for number in number_string:
            number_new.append(number)
            
        return number_new