Leetcode 161 Editing distance of 1 (2022.06.10)

Given two strings s and t , If their edit distance is 1 , Then return to true , Otherwise return to false .

character string s And string t The editing distance between them is equal to 1 There are three possible situations ：

Go to s Insert Just one Character get t
from s Delete in Just one Character get t
stay s of use A different character Replace Just one Character get t

Example 1：

Input : s = “ab”, t = “acb”
Output : true
explain : Can be ‘c’ Insert string s To get t.

Example 2:

Input : s = “cab”, t = “ad”
Output : false
explain : Unable to get 1 Step operation s Turn into t.

Tips :

0 <= s.length, t.length <= 10⁴
s and t It's made up of lowercase letters , Capital letters and numbers

source ： Power button （LeetCode）
link ：https://leetcode.cn/problems/one-edit-distance

Method 1 ： Double pointer

Python3 Submission ：

class Solution:
    def isOneEditDistance(self, s: str, t: str) -> bool:
        i = j = 0
        while(i < len(s) and j < len(t)):
            if s[i] == t[j]:
                i += 1
                j += 1
            else:
                return (s[i+1:] == t[j:]) or (s[i:] == t[j+1:]) or (s[i+1:] == t[j+1:]) 
        return  abs(len(s) - len(t)) == 1