def anagramSolution1(s1, s2):
"""
变位词第一种解法:O(n^2)
:param s1:
:param s2:
:return:
"""
alist = list(s2) # s2字符串转换为列表
pos1 = 0
stillOK = True
while pos1 < len(s1) and stillOK: # 循环s1的每个字符
pos2 = 0
found = False
while pos2 < len(alist) and not found:
if s1[pos1] == alist[pos2]: # 与s2元素逐个对比
found = True
else:
pos2 += 1
if found:
alist[pos2] = None # 找到，打钩
else:
stillOK = False # 未找到，失败
pos1 += 1
return stillOK
print(anagramSolution1('python', 'typhon'))
def anagramSolution2(s1, s2):
"""
变位词第二种解法:O(nlogn)
:param s1:
:param s2:
:return:
"""
alist = list(s1)
blist = list(s2)
alist.sort() # 分别排序
blist.sort()
matches = True
pos1 = 0
while pos1 < len(alist) and matches:
if alist[pos1] == blist[pos1]: # 一一对比
matches = True
else:
matches = False
pos1 += 1
return matches
print(anagramSolution2('abcd', 'dcba'))
def anagramSolution3(s1, s2):
"""
变位词第三种解法，计数比较变位词:O(n)
:param s1:
:param s2:
:return:
"""
c1 = [0] * 26
c2 = [0] * 26
for i in range(len(s1)):
pos = ord(s1[i]) - ord('a') # 分别计数
c1[pos] += 1
for i in range(len(s2)):
pos = ord(s1[i]) - ord('a')
c2[pos] += 1
j = 0
stillOK = True
while j < 26 and stillOK: # 计数器比较
if c1[j] == c2[j]:
j+=1
else:
stillOK = False
return stillOK
print(anagramSolution3('apple', 'pleap'))
