leetcode 面試一次編輯01.05python3

"""字符串有三種編輯操作:插入一個字符、刪除一個字符或者替換一個字符。 給定兩個字符串，編寫一個函數判定它們是否只需要一次(或者零次)編輯。
示例 1:
輸入:
first = "pale"
second = "ple"
輸出: True
示例 2:
輸入:
first = "pales"
second = "pal"
輸出: False
"""
class Solution:
    def get_flag(self,first,second):#獲取兩個字典中不同的個數 eg:建值一個存在一個不存在 value不同
        count = 0
        if len(first)>len(second):
            if second in first:
                return True
        elif len(first)<len(second):
            if first in second:
                return True
        while len(first)>1 or len(second)>1:
            if first[0]==second[0]:
                first = first[1:]
                second = second[1:]
            else:
                count+=1
                first = first[::-1]
                second = second[::-1]
            print("first:",first)
            print("second:",second)
            if count == 2 :
                return False
        return True

    def oneEditAway(self, first: str, second: str) -> bool:
        if abs(len(first)-len(second))>1:
            flag = False
        else :
            flag = self.get_flag(first,second)
        return flag
A = Solution()
print(A.oneEditAway("ab","bc"))