python實(shí)現(xiàn)對變位詞的判斷方法

def anagramSolution1(s1,s2):
 alist = list(s2) # 復(fù)制s2
 pos1 = 0
 stillok = True
 while pos1 < len(s1) and stillok: # 循環(huán)s1的所有字符
  pos2 = 0
  found = False  # 初始化標(biāo)識符
  while pos2 < len(alist) and not found:  # 與s2的字符逐個對比
   if s1[pos1] == alist[pos2]:
    found = True
   else:
    pos2 = pos2 + 1
  if found:
   alist[pos2] = None # 找到對應(yīng)，標(biāo)記
  else:
   stillok = False # 沒有找到，失敗
  pos1 = pos1 + 1
 return stillok
print(anagramSolution1('python','typhon'))

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

 alist1.sort()   # 對字符串進(jìn)行順序排序
 alist2.sort()
 pos = 0
 matches = True
 while pos < len(s1) and matches:
  if alist1[pos] == alist2[pos]:  # 逐個對比
   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函數(shù)返回字符的Unicode編碼，此語句可以將字符串中的字母轉(zhuǎn)換成0-25的數(shù)字
  c1[pos] = c1[pos] + 1  # 實(shí)現(xiàn)計(jì)數(shù)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:   # 計(jì)數(shù)器比較
  if c1[j] == c2[j]:
   j = j + 1
  else:
   stillOK = False
 return stillOK
print(anagramSolution4('python','typhon'))