Python implementation of Leetcode Primary algorithm--string

Python implementation of Leetcode Primary algorithm--string

# invert string def reversestring (s): Return s[::-1]# reverse number def reverse (x): if x < 0:flag =-2 * * result =         -1 * INT (str (x) [1:][::-1]) if result < Flag:return 0 Else:return result Else:            Flag = 2 * * 31-1 result = Int (str (x) [:: -1]) If result > Flag:return 0 Else: Return result# the first unique character in a string def firstuniqchar (s): D = {} for I in range (len (s)): d[s[i]] = D.get (s[ I], 0) + 1 for I in range (len (s)): if d[s[i]] = = 1:return I return-1# valid letter ectopic word def isanagram (s, t        ): If Len (t)! = Len (s): return False If Len (set (t))! = Len (set (s)): return False for ex in Set (T):    If S.count (ex)! = T.count (ex): Return False return true# verify palindrome string def ispalindrome (s): Import re  s = s.lower () NewS = Re.sub (R ' [^a-za-z0-9] ', "", s) if news[::-1] = = News:return True Else:return false# string to Integer (atoi) defMyatoi (str): Import re if Re.match (' \s+ ', str)! = none:a, B = re.match (' \s+ ', str). span () str = str[b :] if str = = ": return 0 flag = True if str[0] = = '-': str = str[1:] Flag = False elif s Tr[0] = = ' + ': str = str[1:] If Re.match (' \d+ ', str)! = none:a, B = re.match (' \d+ ', str). Span () St            r = str[a:b] If flag = = True:if int (str) > 2 * * 31-1: Return 2 * * 31-1 return int (str) else:if-1 * INT (str) <-2 * * 31:return-2 * * return -1 * INT (str) Else:return 0# implements the StrStr () function.        def strStr (haystack, needle): If Len (needle) = = 0:return 0 for i in range (len (haystack)-Len (needle) + 1):   If haystack[i:i + len (needle)] = = Needle:return I return-1# number and say Def countandsay (n): keystr = ' 1 ' # starting from 1 for I in Range (n-1): # loop n times newstr = "" # store new characters strlist = []  # Loop A string down to get the number of each character sList = [] # Number of identical characters flag = True for j in Range (Len (KEYSTR)-1): # loop character length minus A flag indicates that the default last character differs from the previous one, Slist.append (Keystr[j]) slist.append (1) if keystr[j] = = Keystr                [j + 1]: # If the character of the current position is the same as the character at the next position slist[1] + = 1 flag = False Else: Strlist.append (sList) # different words store the case of the previous character in the list sList = [] flag = True If flag: # if the last                Character is different from the previous character, the character condition is added to Strlist.append ([keystr[-1], 1]) else:if sList! = []: # Last string of the same character added to the list Strlist.append (sList) for K in range (len (strlist)): # takes the characters of the list out sequentially newstr = newstr + ' + S TR (strlist[k][1]) Newstr = newstr + ' + strlist[k][0] keystr = newstr return keystr# longest common prefix def longe Stcommonprefix (STRs): if STRs = = []: return ' Lenkey = Len (strs[0]) key = 0 Longest = 0 for i in RA      Nge (1, Len (STRs)):  If Len (Strs[i]) < Lenkey:key = I lenkey = Len (strs[i]) for I in Range (len (Strs[key])):        Flag = True for j in Range (Len (STRs)): if strs[j][i]! = Strs[key][i]: flag = False If flag:longest + = 1 else:break return strs[key][0:longest]

Python implementation of the Leetcode primary algorithm--string