遇到一个算法题，接龙的单词，求大神相助！！！

[无名鼠辈]

在公司Mala-Offline,人们喜欢玩一个游戏叫做接龙的单词。玩这个游戏,他们定义一组
单词的字典。然后选择任意两个单词从字典表示开始和结束,我们的目标是找到
从开始到结束的所有最短序列(s),比如:
•每次只能改变一个字母
•每个中间单词必须存在于字典

给出：
start = "hlt"
end = "cag"
dict = ["hat","dat","dag","lat","lag"]

返回：
[
["hlt","hat","dat","dag","cag"],
["hlt","hat","lat","lag","cag"]
]
不知道算法应该如何写？求代码。。。

[飏飏一蝶]

感觉可以转换成图的最短路径算法之类的

[无名鼠辈]

能不能有具体点的代码呢？

[ITjavaman]

语文没学好，表示没看懂

[无名鼠辈]

引用 1 楼 飏飏一蝶的回复:

感觉可以转换成图的最短路径算法之类的

能不能有具体点的代码呢～

[qq_36425552]

package 乱七八糟程序; import java.util.ArrayList; import java.util.Iterator; import java.util.Scanner; import java.util.TreeSet; /** * Created by LXM on 2017-05-21. */ public class ProgramExam { static ArrayList<ArrayList<EngWord>> finalyRes = new ArrayList<>(); //存放最后所有的接龙 static TreeSet<String> finalyRes2 = new TreeSet<>(); //最长的可能有多条，用这个来存放 public static void main(String[] args) { System.out.println("please input some Word："); Scanner sc = new Scanner(System.in); String read = sc.nextLine(); String[] eachPath = read.split("\\s+"); //递归 change(eachPath, 0, eachPath.length - 1); //找最长 int max = 0; for (int i = 0; i < finalyRes.size(); i++) { if (finalyRes.get(i).size() > max) max = finalyRes.get(i).size(); } //输出所有最长 for (int i = 0; i < finalyRes.size(); i++) { String tmp = ""; if (finalyRes.get(i).size() == max) { Iterator<EngWord> iterator = finalyRes.get(i).iterator(); while (iterator.hasNext()) { EngWord eng = iterator.next(); tmp += eng.getExp() + " "; } if (!"".equals(tmp)) finalyRes2.add(tmp); } } Iterator<String> iterator = finalyRes2.iterator(); while (iterator.hasNext()) { System.out.println(iterator.next()); } } public static void change(String[] str, int first, int ends) { if (first == ends) { ArrayList<EngWord> arr = new ArrayList<>(); ArrayList<EngWord> res = new ArrayList<>(); for (int i = 0; i < str.length; i++) { char[] tmp = str[i].toCharArray(); char start = tmp[0]; char end = tmp[tmp.length - 1]; EngWord engWord = new EngWord(str[i], start, end); arr.add(engWord); } boolean flag = true; //be eh hi il //轮询 for (int i = 0; i < arr.size(); i++) { res.clear(); EngWord baseWord = arr.get(i); res.add(baseWord); for (int k = 0; k < arr.size(); k++) { for (int j = 0; j < arr.size(); j++) { if (i == j) continue; EngWord secondWord = arr.get(j); if (secondWord.getStart() == res.get(res.size() - 1).getEnd()) { if (!res.contains(secondWord)) { res.add(secondWord); j = 0; } } else { continue; } } } //只有一个的话就不接了 if (res.size() == 1) { continue; } //存下每一个可以接龙的 ArrayList<EngWord> copyArr = (ArrayList<EngWord>) res.clone(); finalyRes.add(copyArr); } } for (int i = first; i <= ends; i++) { swap(str, i, first); change(str, first + 1, ends); //固定好当前一位，继续排列后面的 swap(str, i, first); } } private static void swap(String[] str, int i, int first) { String tmp; tmp = str[first]; str[first] = str[i]; str[i] = tmp; } } class EngWord { String exp; char start; char end; public EngWord(String exp, char s, char e) { this.exp = exp; this.start = s; this.end = e; } public EngWord() { } public String getExp() { return exp; } public void setExp(String exp) { this.exp = exp; } public char getStart() { return start; } public void setStart(char start) { this.start = start; } public char getEnd() { return end; } public void setEnd(char end) { this.end = end; } }