[LeetCode][JavaScript]Word Ladder

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

Word Ladder

Given two words (beginWord and endWord), and a dictionary, find the length of shortest transformation sequence from beginWord to endWord, such that:

1. Only one letter can be 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 is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Note:

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

这题挂了无数次啊，一开始错把入参当做Array，其实是个Set，之后就一直超时...

bfs的思路是没错的，但首先不能构造图，复杂度O(n^2) 肯定不行。

然后尝试直接在Set里找，找到一个删除一个，也是超时。

最后只能换成了现在的方式。

这道题test case有特殊性，dict里的单词不会特别长，但是数据量很大。

于是最终改成在bfs队列中出队一个元素，一个个地用a-z替换每个字母，然后去跟字典比较。

 1 /**
 2  * @param {string} beginWord
 3  * @param {string} endWord
 4  * @param {set<string>} wordDict
 5  * @return {number}
 6  */
 7 var ladderLength = function(beginWord, endWord, wordDict) {
 8     var queue = [];
 9     var i = 0;
10     queue.push(beginWord);
11     wordDict.delete(beginWord);
12     if(oneCharDiff(beginWord, endWord) && wordDict.has(endWord)){
13         return 2;
14     }else{
15         return bfs();
16     }
17
18     function bfs(){
19         var depth = 1;
20         while(queue.length > 0){
21             depth++;
22             var count = queue.length;
23             while(count--){
24                 var curr = queue.shift();
25                 if(oneCharDiff(curr, endWord) && curr !== beginWord){
26                     return depth;
27                 }
28                 var needRemove = [];
29                 for(var i = 0; i < curr.length; i++){
30                     for(var j = ‘a‘.charCodeAt(); j <= ‘z‘.charCodeAt(); j++){
31                         var testMatch = curr;
32                         var ch = String.fromCharCode(j);
33                         if(testMatch[i] !== ch){
34                             testMatch = replaceChat(testMatch, i, ch);
35                         }
36                         if(wordDict.has(testMatch)){
37                             queue.push(testMatch);
38                             wordDict.delete(testMatch);
39                         }
40                     }
41                 }
42             }
43         }
44
45         return 0;
46     }
47     function isStrInArr(str, arr){
48         for(var i in arr){
49             if(str === arr[i]){
50                 return true;
51             }
52         }
53         return false;
54     }
55     function replaceChat(source, pos, newChar){
56         var sFrontPart = source.substr(0, pos);
57         var sTailPart = source.substr(pos + 1, source.length);
58         return sFrontPart + newChar + sTailPart;
59     }
60     function oneCharDiff(a, b){
61         if(a.length !== b.length){
62             return false;
63         }
64         var count = 0;
65         for(var i = 0; i < a.length; i++){
66             if(a[i].toLowerCase() !== b[i].toLowerCase()){
67                 count++;
68             }
69             if(count >= 2){
70                 return false;
71             }
72         }
73         if(count === 1){
74             return true;
75         }else{
76             return false;
77         }
78     }
79 };