python 一次编辑

字符串有三种编辑操作:插入一个字符、删除一个字符或者替换一个字符。 给定两个字符串，编写一个函数判定它们是否只需要一次(或者零次)编辑。

 

示例 1:

输入: 
first = "pale"
second = "ple"
输出: True
 

示例 2:

输入: 
first = "pales"
second = "pal"
输出: False

class Solution:
    """
    解题思路：
    1.先判断两个字符串长度是否相等 
    2.判断两个字符串长度的差值石佛等于1
        1.找出字符串长度比较长的哪一个,当有值不相等时,去给长度长的那个字符串的索引值+1
    """
    def oneEditAway(self, first: str, second: str) -> bool:
        first_len = len(first)
        second_len = len(second)
        different_num = 0
        if first_len == second_len:
            if first_len == 1:
                return True
            for i in range(first_len):
                if first[i] != second[i]:
                    different_num += 1
            return False if different_num > 1 else True
        elif abs(first_len-second_len) == 1:
            if first_len == 0 or second_len == 0:
                return True
            first_index = 0
            second_index = 0
            # 当第一个大于第二个的时候，让其交换位置
            if first_len > second_len:
                first, second = second, first
                first_len = len(first)
                second_len = len(second)
            while first_len > first_index and second_len > second_index: 
                if first[first_index] != second[second_index]:
                    different_num += 1
                    second_index += 1 
                else:
                    first_index += 1
                    second_index += 1
            return False if different_num > 1 else True 

        else:
            return False