判断两个字符串是否接近

Question

我正在尝试制作一个程序，比较word1字符串与word2字符串只发生一次

class Solution:
def closeStrings(self, word1: str, word2: str) -> bool:
word1 = [x.strip() for x in word1]
word2 = [x.strip() for x in word2]
update = False
for x in word1:
if(x in word2):
update = True
if(type(x) is str):

a = word1.index(x)
b = word2.index(x)
word1[a]=''
word2[b]=''
else:
update = False
else:
update = False
break

return update
print(Solution.closeStrings(Solution,word1='a',word2='aa'))

输入

word1 = 'a',word2 ='aa'

预期Output = False

实际Output = True

Answer 1

1. print(Solution.closeStrings(Solution,word1='a',word2='aa'))

   创建一个类是为了能够创建它的实例。这样，您就不需要将Solution作为self参数传递。

2. word1 = [x.strip() for x in word1]

   看起来你想要删除空格。但是你会得到一个字符串列表，其中的空格为空字符串。那不是你想要的。查看

   的输出print([x.strip() for x in "Hello world"])

3. 你的算法太复杂了。

   您可以简单地计算word2:

   中每个字符的出现次数。

class Solution:
def closeStrings(self, word1: str, word2: str) -> bool:
for x in word1:
if word2.count(x) != word1.count(x): return False
return True

s = Solution()
print(s.closeStrings(word1='a',word2='aa'))
print(s.closeStrings(word1='abcb',word2='bcab'))

Answer 2

扩展到其他更多的解决方案的答案@Thomas Weller很好地解释了他

class Solution:
def closeStrings(self, word1: str, word2: str) -> bool:
for i in word1:
if i not in word2:
return False
for i in word2:
if i not in word1:
return False
return True
def closeStrings2(self, word1: str, word2: str) -> bool:
if len(word1) != len(word2):
return False
if set(word1) != set(word2):
return False
return True
def closeStrings3(self, word1: str, word2: str) -> bool:
if len(word1) != len(word2):
return False
if sorted(word1) != sorted(word2):
return False
return True
print(Solution().closeStrings(word1="cabbba", word2="abbccc"))
print(Solution().closeStrings3(word1="cabbba", word2="aabbss"))
print(Solution().closeStrings3(word1="cabbba", word2="aabbss"))