Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:
- Only one letter can be changed at a time
- 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.
题解：
这道题是套用BFS同时也利用BFS能寻找最短路径的特性来解决问题。
把每个单词作为一个node进行BFS。当取得当前字符串时，对他的每一位字符进行从a~z的替换，如果在字典里面，就入队，并将下层count++，并且为了避免环路，需把在字典里检测到的单词从字典里删除。这样对于当前字符串的每一位字符安装a~z替换后，在queue中的单词就作为下一层需要遍历的单词了。
正因为BFS能够把一层所有可能性都遍历了，所以就保证了一旦找到一个单词equals（end），那么return的路径肯定是最短的。
像给的例子 start = hit，end = cog，dict = [hot, dot, dog, lot, log]
按照上述解题思路的走法就是：
level = 1 hit dict = [hot, dot, dog, lot, log]
ait bit cit ... xit yit zit ， hat hbt hct ... hot ... hxt hyt hzt ， hia hib hic ... hix hiy hiz
level = 2 hot dict = [dot, dog, lot, log]
aot bot cot dot ... lot ... xot yot zot，hat hbt hct ... hxt hyt hzt，hoa hob hoc ... hox hoy hoz
level = 3 dot lot dict = [dog log]
aot bot ... yot zot，dat dbt ...dyt dzt，doa dob ... dog .. doy doz，
aot bot ... yot zot，lat lbt ... lyt lzt，loa lob ... log... loy loz
level = 4 dog log dict = []
aog bog cog
level = 5 cog dict = []
代码如下：
2 if(start==null || end==null || start.length()==0 || end.length()==0 || start.length()!=end.length())
3 return 0;
4
5 LinkedList<String> wordQueue = new LinkedList<String>();
6 int level = 1; //the start string already count for 1
7 int curnum = 1;//the candidate num on current level
8 int nextnum = 0;//counter for next level
9
10 wordQueue.add(start);
11
12 while(!wordQueue.isEmpty()){
13 String word = wordQueue.poll();
14 curnum--;
15
16 for(int i = 0; i < word.length(); i++){
17 char[] wordunit = word.toCharArray();
18 for(char j = 'a'; j <= 'z'; j++){
19 wordunit[i] = j;
20 String temp = new String(wordunit);
21
22 if(temp.equals(end))
23 return level+1;
24 if(dict.contains(temp)){
25 wordQueue.add(temp);
26 nextnum++;
27 dict.remove(temp);
28 }
29 }
30 }
31
32 if(curnum == 0){
33 curnum = nextnum;
34 nextnum = 0;
35 level++;
36 }
37 }
38
39 return 0;
40 }