python变位词判断 python实现对变位词的判断方法

def anagramSolution1(s1,s2):
 alist = list(s2) # 复制s2
 pos1 = 0
 stillok = True
 while pos1 < len(s1) and stillok: # 循环s1的所有字符
  pos2 = 0
  found = False  # 初始化标识符
  while pos2 < len(alist) and not found:  # 与s2的字符逐个对比
   if s1[pos1] == alist[pos2]:
    found = True
   else:
    pos2 = pos2 + 1
  if found:
   alist[pos2] = None # 找到对应，标记
  else:
   stillok = False # 没有找到，失败
  pos1 = pos1 + 1
 return stillok
print(anagramSolution1('python','typhon'))

def anagramSolution2(s1,s2):
 alist1 = list(s1)
 alist2 = list(s2)

 alist1.sort()   # 对字符串进行顺序排序
 alist2.sort()
 pos = 0
 matches = True
 while pos < len(s1) and matches:
  if alist1[pos] == alist2[pos]:  # 逐个对比
   pos = pos + 1
  else:
   matches = False
 return matches
print(anagramSolution2('python','typhon'))

def anagramSolution4(s1,s2):
 c1 = [0] * 26
 c2 = [0] * 26
 for i in range(len(s1)):
  pos = ord(s1[i]) - ord('a')  # ord函数返回字符的Unicode编码，此语句可以将字符串中的字母转换成0-25的数字
  c1[pos] = c1[pos] + 1  # 实现计数器
 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:   # 计数器比较
  if c1[j] == c2[j]:
   j = j + 1
  else:
   stillOK = False
 return stillOK
print(anagramSolution4('python','typhon'))