python判断是否为变位词

python判断是否为变位词

class AnagramDetection:

# 先对两个字符串进行list化

# 对字符串对应的两个list进行排序

# 依次比较字符是否匹配

def anagramSolution1(self, s1, s2):

alist1 = list(s1)

alist2 = list(s2)

alist1.sort()

alist2.sort()

pos = 0

matches = True

while pos < len(s1) and matches:

if alist1[pos] == alist2[pos]:

pos = pos + 1

else:

matches = False

return matches

# 首先生成两个26个字母的list

# 计算每个字母出现的次数并存入到相应的list中

# 比较两个list是否相同

def anagramSolution2(self, s1, s2):

c1 = [0] * 26

c2 = [0] * 26

for i in range(len(s1)):

pos = ord(s1[i]) - ord('a')

c1[pos] = c1[pos] + 1

for i in range(len(s2)):

pos = ord(s2[i]) - ord('a')

c2[pos] = c2[pos] + 1

j = 0

stillOK = True

while j < 26 and stillOK:

if c1[j] == c2[j]:

j = j + 1

else:

stillOK = False

return stillOK

# 首先将两个字符串list化

# 将两个list中的字符生成两个set

# 比较两个set, 如果不相等直接返回false

# 如果两个set相等, 比较每个set中字符在相应list中的个数, 个数不同返回false

def anagramSolution3(self, s1, s2):

alist1 = list(s1)

alist2 = list(s2)

aset1 = set(alist1)

aset2 = set(alist2)

if aset1 != aset2:

return False

else:

for ch in aset1:

unt(ch) != unt(ch):

return False

return True

s1 = 'abcde'

s2 = 'acbde'

test = AnagramDetection()

print(test.anagramSolution1(s1, s2))

print(test.anagramSolution2(s1, s2))

print(test.anagramSolution3(s1, s2))