[Leetcode] [JavaScript] Word Ladder

https://leetcode.com/problems/word-ladder/

Word Ladder

Given words (beginword and Endword), and a dictionary, find the length of shortest transformation se Quence from Beginword to Endword, such that:

1. One letter can is changed at a time
2. Each intermediate word must exist in the dictionary

For example,

Given:
Start ="hit"
End ="cog"
dict =["hot","dot","dog","lot","log"]

As one shortest transformation "hit" -> "hot" -> "dot" -> "dog" -> "cog" is,
return its length 5 .

Note:

• Return 0 If there is no such transformation sequence.
• All words has the same length.
• All words contain only lowercase alphabetic characters.

Sure enough JS is not suitable to do the problem, the lack of a variety of common methods.

This question hung countless times ah, at the beginning of the mistake into the parameter as an array, is actually a set, and then has been timed out ...

The idea of BFS is correct, but first can not construct the diagram, the Complexity O (n^2) certainly not.

Then try to find it directly in set, find a delete, and also timeout.

In the end, it can only be replaced by the present way.

The test case has a special character, dict words are not particularly long, but the amount of data is very large.

It eventually changed to an element in the BFS queue, replacing each letter with a-Z and then comparing it to the dictionary.

1 /**2 * @param {string} Beginword3 * @param {string} Endword4 * @param {set<string>} worddict5 * @return {number}6  */7 varLadderlength =function(Beginword, Endword, worddict) {8     varQueue = [];9     vari = 0;Ten Queue.push (Beginword); OneWorddict.Delete(Beginword); A     if(Onechardiff (Beginword, Endword) &&Worddict.has (Endword)) { -         return2; -}Else{ the         returnBFS (); -     } -      -     functionBFs () { +         vardepth = 1; -          while(Queue.length > 0){ +depth++; A             varCount =queue.length; at              while(count--){ -                 varCurr =Queue.shift (); -                 if(Onechardiff (Curr, Endword) && Curr!==Beginword) { -                     returndepth; -                 } -                 varNeedremove = []; in                  for(vari = 0; i < curr.length; i++){ -                      for(varj = ' A '. charCodeAt (); J <= ' Z '. charCodeAt (); J + +){ to                         varTestmatch =Curr; +                         varCH =String.fromCharCode (j); -                         if(Testmatch[i]!==ch) { theTestmatch =Replacechat (Testmatch, I, ch); *                         } $                         if(Worddict.has (Testmatch)) {Panax Notoginseng Queue.push (testmatch); -Worddict.Delete(Testmatch); the                         } +                     } A                 } the             } +         }         -          $         return0; $     } -     functionIsstrinarr (str, arr) { -          for(varIincharr) { the             if(str = = =Arr[i]) { -                 return true;Wuyi             } the         } -         return false; Wu     } -     functionreplacechat (source, POS, Newchar) { About         varSfrontpart = source.substr (0, POS);  $         varStailpart = SOURCE.SUBSTR (pos + 1, source.length);  -         returnSfrontpart + Newchar +Stailpart;  -     }   -     functionOnechardiff (A, b) { A         if(A.length!==b.length) { +             return false; the         } -         varCount = 0; $          for(vari = 0; i < a.length; i++){ the             if(A[i].tolowercase ()!==b[i].tolowercase ()) { thecount++; the             } the             if(Count >= 2){ -                 return false; in             } the         } the         if(Count = = 1){ About             return true; the}Else{ the             return false; the         } +     } -};

[Leetcode] [JavaScript] Word Ladder