class Solution(object):
def isOneEditDistance(self, word1, word2):
l1=len(word1)
l2=len(word2)
if l1>l2:
return self.isOneEditDistance(word2,word1)
if l2-l1>1:
return False
i=0
while i<l1 and word1[i]==word2[i]:
i+=1
if l1==l2:
return word1[i+1:]==word2[i+1:]
else:
return word1[i:]==word2[i+1:]
161. One Edit Distance
最后编辑于 :
©著作权归作者所有,转载或内容合作请联系作者
平台声明:文章内容(如有图片或视频亦包括在内)由作者上传并发布,文章内容仅代表作者本人观点,简书系信息发布平台,仅提供信息存储服务。
平台声明:文章内容(如有图片或视频亦包括在内)由作者上传并发布,文章内容仅代表作者本人观点,简书系信息发布平台,仅提供信息存储服务。
推荐阅读更多精彩内容
- Given two strings S and T, determine if they are both one...
- Given two strings S and T, determine if they are both one...
- Given two strings S and T, determine if they are both one...
- Given two strings S and T, determine if they are both one...