leetcode:Word Ladder II

Posted 2020-10-06

　　题目大意：给出一个字符串列表list，一个源字符串，一个目标字符串，且这些字符串拥有相同长度。要求每次最多变换源字符串中一个字符，且每次变换的结果都应该在list中有相同的字符串。求所有从源字符串到目标字符串变换次数最少的策略。　　

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　　相当简单的题目，可以将每个字符串看作一个顶点，而若两个字符串之间只有一个字符不同，那么就认为这两个字符串之间有双向边。这样就构建了一幅图，之后从源字符串对应的顶点出发，去寻找到达目标字符串的方案，就相当于找从源顶点出发到目标顶点的最短路径。这个过程利用广度优先搜索即可实现，搜索的时间复杂度最多为O(V)+O(E)<=O(V^2)。

　　时间复杂度是非多项式的，因为题目还要求给出所有最短的路径。这样可能的总共的路径数目可以达到O(2^n)级别。因此其实这道题目并不好，之所以能通过也只是因为数据不够变态。

　

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  1 package cn.dalt.leetcode;
  2 
  3 import java.util.*;
  4 
  5 /**
  6  * Created by Administrator on 2017/9/8.
  7  */
  8 public class WordLadderII {
  9     public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {
 10         if (beginWord.equals(endWord)) {
 11             return Arrays.asList(Arrays.asList(beginWord));
 12         }
 13 
 14         boolean pass = false;
 15         for (String word : wordList) {
 16             if (word.equals(endWord)) {
 17                 pass = true;
 18                 break;
 19             }
 20         }
 21         if (!pass) {
 22             return Collections.emptyList();
 23         }
 24 
 25         int n = wordList.size();
 26         int m = beginWord.length();
 27 
 28         Node[] node = new Node[n + 2];
 29         for (int i = 0, bound = wordList.size(); i < bound; i++) {
 30             node[i] = new Node();
 31             node[i].signature = wordList.get(i);
 32             node[i].flag = Integer.MAX_VALUE;
 33         }
 34 
 35         node[n] = new Node();
 36         node[n].signature = beginWord;
 37         node[n + 1] = new Node();
 38         node[n + 1].signature = endWord;
 39         Node src = node[n];
 40         Node dst = node[n + 1];
 41         src.flag = 0;
 42         dst.flag = Integer.MAX_VALUE;
 43 
 44         //Build graph
 45         for (int i = 0, bound = n + 2; i < bound; i++) {
 46             for (int j = i + 1; j < bound; j++) {
 47                 if (differForOneLetter(node[i].signature, node[j].signature)) {
 48                     node[i].nearTo.add(node[j]);
 49                     node[j].nearTo.add(node[i]);
 50                 }
 51             }
 52         }
 53 
 54         //BFS
 55         Deque<Node> queue = new LinkedList<>();
 56         queue.add(src);
 57         while (!queue.isEmpty()) {
 58             Node front = queue.pollFirst();
 59             int nextFlag = front.flag + 1;
 60             for (Node near : front.nearTo) {
 61                 if (near.flag == nextFlag) {
 62                     near.parent.add(front);
 63                 } else if (near.flag > nextFlag) {
 64                     near.flag = nextFlag;
 65                     near.parent.add(front);
 66                     queue.addLast(near);
 67                 }
 68             }
 69         }
 70 
 71         List<List<String>> result = new LinkedList<>();
 72         if (dst.flag == Integer.MAX_VALUE) {
 73             return Collections.emptyList();
 74         }
 75         dfs(new String[dst.flag + 1], dst.flag, dst, result);
 76         return result;
 77     }
 78 
 79     public void dfs(String[] route, int index, Node current, List<List<String>> routes) {
 80         route[index] = current.signature;
 81         if (index == 0) {
 82             routes.add(Arrays.asList(route.clone()));
 83             return;
 84         }
 85         for (Node parent : current.parent) {
 86             dfs(route, index - 1, parent, routes);
 87         }
 88     }
 89 
 90     public boolean differForOneLetter(String a, String b) {
 91         int differNum = 0;
 92         for (int i = 0, bound = a.length(); i < bound; i++) {
 93             differNum += a.charAt(i) == b.charAt(i) ? 0 : 1;
 94         }
 95         return differNum == 1;
 96     }
 97 
 98     private static class Node {
 99         String signature;
100         int flag;
101         List<Node> parent = new LinkedList<>();
102         List<Node> nearTo = new LinkedList<>();
103 
104         @Override
105         public String toString() {
106             return signature + ":" + flag;
107         }
108     }
109 }