Lint Code 120. 单词接龙

虽然知道是使用bfs求树节点之间的最短路径问题但是还是码不出来，看的答案，失败，具体思路和二叉树的层次遍历近似

import org.junit.Test;import java.util.*;public class LadderLength {
    /*** @param start: a string* @param end: a string* @param dict: a set of string* @return: An integer* <p>* 120. 单词接龙* 给出两个单词（start和end）和一个字典，找到从start到end的最短转换序列* <p>* 比如：* <p>* 1.每次只能改变一个字母。* 2.变换过程中的中间单词必须在字典中出现。* 样例* 给出数据如下：* <p>* start = "hit"* <p>* end = "cog"* <p>* dict = ["hot","dot","dog","lot","log"]* <p>* 一个最短的变换序列是 "hit" -> "hot" -> "dot" -> "dog" -> "cog"，* <p>* 返回它的长度 5* <p>* 注意事项* 如果没有转换序列则返回0。* 所有单词具有相同的长度。* 所有单词都只包含小写字母。* <p>* 求图的最短路径使用bfs*/public int ladderLength(String start, String end, Set<String> dict) {// write your code hereif (dict == null) {return 0;}if (start.equals(end)) {return 1;}//hashSet用于过滤已经使用的dict中的元素，防止再走一遍HashSet<String> hashSet = new HashSet<>();//queue中放每一层的节点LinkedList<String> linkedList = new LinkedList<>();hashSet.add(start);linkedList.add(start);//可能dict中没有end，但是会有end的上一步的worddict.add(end);//下面开始遍历一层中所有的next，想象成一颗二叉树的层次遍历int result = 1;while (!linkedList.isEmpty()) {//遍历的层数就是路径的长度result++;int layerSize = linkedList.size();for (int i = 0; i < layerSize; i++) {String poll = linkedList.poll();for (String nextWord : getNextWords(poll,dict)) {if (hashSet.contains(nextWord)) {continue;}if (nextWord.equals(end)) {return result;}hashSet.add(nextWord);linkedList.add(nextWord);}}}return 0;}private String replace(String s, int index, char c) {char[] chars = s.toCharArray();chars[index] = c;return new String(chars);}private ArrayList<String> getNextWords(String word, Set<String> dict) {char[] chars = word.toCharArray();ArrayList<String> nextWords = new ArrayList<>();for (int j = 0; j < chars.length; j++) {for (char i = 'a'; i <= 'z'; i++) {if (chars[j] == i) {continue;}String nextString = replace(word, j, i);if (dict.contains(nextString)) {nextWords.add(nextString);}}}return nextWords;}@Testpublic void testGetNextWords() {String word = "chen";Set<String> dict = new HashSet<>();dict.add("chea");dict.add("cheb");dict.add("chec");dict.add("chan");dict.add("chbn");dict.add("chba");dict.add("chca");dict.add("cdca");// ArrayList<String> nextWords = getNextWords(word, dict);
// System.out.println(Arrays.toString(nextWords.toArray()));System.out.println(ladderLength(word,"cdca",dict));}}