题解 | #编辑距离为一#

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param s string字符串 
# @param t string字符串 
# @return bool布尔型
#
class Solution:
    def editdistance(self , s: str, t: str) -> bool:
        # 以下注释的写法会部分超时
        # def inner(s1,s2):
        #     dp=[[0 for j in range(len(s2)+1)] for i in range(len(s1)+1)]
        #     dp[0]=[j for j in range(len(s2)+1)]
        #     for i in range(len(s2)+1):
        #         dp[i][0]=i
        #     for i in range(1,len(s1)+1):
        #         for j in range(1,len(s2)+1):
        #             if s1[i-1]==s2[j-1]:
        #                 dp[i][j]=dp[i-1][j-1]
        #             else:
        #                 dp[i][j]=min(dp[i-1][j],dp[i][j-1],dp[i-1][j-1])+1
        #     return dp[len(s1)][len(s2)]
        # if inner(s,t)==1:
        #     return True
        # else:
        #     return False
        L1,L2=len(s),len(t)
        if abs(L2-L1)>1:
            return False
        elif L1==L2:
            cou=0
            for i in range(L1):
                if s[i]!=t[i]:
                    cou+=1
            if cou==1:
                return True
            else:
                return False
        else:
            s1,s2=(s,t) if L1<=L2 else (t,s)
            for i in s1:
                if i not in s2:
                    return False
            return True