[LeetCode] #66 Plus One (Easy)
LeetCode 第 66 題 Plus One，難度 Easy
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
- 把輸入的陣列數字一個一個串成字串
- 字串轉換成數字，然後 +1
- 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
