leetcode--Word Solitaire (Python)

Use dictionaries to reduce the complexity of lookups. Using list will time out.

1 classSolution:2 3     defnextwordslist (self, Word, worddict):4Res_list = []5          forIinchRange (len (word)):6              forJinchstring.ascii_lowercase:7New_word =list (word)8                 ifJ! =Word[i]:9New_word[i] =JTenNew_word ="'. Join (New_word) One                     ifNew_wordinchworddict: A res_list.append (New_word) -                         delWorddict[new_word] -         returnres_list the  -  -     defBFS (self, Beginword, Endword, worddict): -         #returns an int +Queue = [] -Queue.append ([Beginword, 1]) +          whileQueue: AWord, step = queue[0][0], queue[0][1] at queue.pop (0) -             ifWord = = Endword:returnStep -             #get the next time you change a word, get a list of words -Nextwords =self.nextwordslist (Word, worddict) -              forJinchnextwords: -Queue.append ([J, Step+1]) in         return0 -     defladderlength (self, Beginword, Endword, wordList): to         """ + : Type Beginword:str - : Type Endword:str the : Type WORDLIST:LIST[STR] * : Rtype:int $         """Panax Notoginseng         ifBeginwordinchWordList:wordList.remove (Beginword) -Worddict = {} the          forWinchWORDLIST:WORDDICT[W] = 1 +         returnSelf.bfs (Beginword, Endword, worddict)

leetcode--Word Solitaire (Python)