LeetCode007:WordLadder

参考：http://www.cnblogs.com/springfor/p/3893499.html

题目：

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, 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.

 

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按照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 = []

 

package com.abuge;import java.util.LinkedList;import java.util.Set;import java.util.TreeSet;import org.junit.Test;/** * 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 timeEach intermediate word must exist in the dictionaryFor 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 * @author AbuGe * */public class Solution {@Testpublic void test(){String start = "hit";String end = "cog";Set<String> dict = new TreeSet<String>();dict.add("hot");dict.add("dot");dict.add("dog");dict.add("lot");dict.add("log");System.out.println(new Solution().laddderLength(start, end, dict));}/** * 思路：（没递归进一层，就去除字典中的相关词，从而避免重复，对于同一层的候选词，谁先到达目的单词谁先返回，则得到最短路径） * 1、使用LinkedList队列维护符合要求的字符串 * 2、通过BFS搜索符合要求的路径 * @param start * @param end * @param dict * @return */public int laddderLength(String start, String end, Set<String> dict){//单词为空，或者长度不等或者长度为零的情况if(start == null || end == null || start.length() != end.length() || start.length() == 0 || end.length() == 0)return 0;//创建一个LinkedList维护单词队列LinkedList<String> wordQueue = new LinkedList<String>();//将起始单词加入队列wordQueue.add(start);//定义相关变量int len = 1;//记录路径长度int curNum = 1;//记录当前层次的单词数int nextNum = 0;//记录下一层的候选单词数//递归调用while(!wordQueue.isEmpty()){String word =  wordQueue.poll();curNum--;for(int i = 0; i < word.length(); i++){char[] wordArray = word.toCharArray();for(char j = 'a'; j < 'z'; j++){wordArray[i] = j;String tmp = new String(wordArray);if(tmp.equals(end))return len + 1;if(dict.contains(tmp)){wordQueue.add(tmp);nextNum++;dict.remove(tmp);}}}if(curNum == 0){curNum = nextNum;nextNum = 0;len++;}}return 0;    }}