120.单词接龙

描述

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

比如

每次只能改变一个字母。
变换过程中的中间单词必须在字典中出现。

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。

样例

给出数据如下：
start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]
一个最短的变换序列是 "hit" -> "hot" -> "dot" -> "dog" -> "cog"，
返回它的长度 5

代码

1. version: LintCode ( Set )

public class Solution {
    public int ladderLength(String start, String end, Set dict) {
        if (dict == null) {
            return 0;
        }

        if (start.equals(end)) {
            return 1;
        }
        
        dict.add(start);
        dict.add(end);

        HashSet hash = new HashSet();
        Queue queue = new LinkedList();
        queue.offer(start);
        hash.add(start);
        
        int length = 1;
        while(!queue.isEmpty()) {
            length++;
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                String word = queue.poll();
                for (String nextWord: getNextWords(word, dict)) {
                    if (hash.contains(nextWord)) {
                        continue;
                    }
                    if (nextWord.equals(end)) {
                        return length;
                    }
                    
                    hash.add(nextWord);
                    queue.offer(nextWord);
                }
            }
        }
        return 0;
    }

    // replace character of a string at given index to a given character
    // return a new string
    private String replace(String s, int index, char c) {
        char[] chars = s.toCharArray();
        chars[index] = c;
        return new String(chars);
    }
    
    // get connections with given word.
    // for example, given word = 'hot', dict = {'hot', 'hit', 'hog'}
    // it will return ['hit', 'hog']
    private ArrayList getNextWords(String word, Set dict) {
        ArrayList nextWords = new ArrayList();
        for (char c = 'a'; c <= 'z'; c++) {
            for (int i = 0; i < word.length(); i++) {
                if (c == word.charAt(i)) {
                    continue;
                }
                String nextWord = replace(word, i, c);
                if (dict.contains(nextWord)) {
                    nextWords.add(nextWord);
                }
            }
        }
        return nextWords;
    }
}

2. version: LeetCode

public class Solution {
    public int ladderLength(String start, String end, List wordList) {
        Set dict = new HashSet<>();
        for (String word : wordList) {
            dict.add(word);
        }
        
        if (start.equals(end)) {
            return 1;
        }
        
        HashSet hash = new HashSet();
        Queue queue = new LinkedList();
        queue.offer(start);
        hash.add(start);
        
        int length = 1;
        while (!queue.isEmpty()) {
            length++;
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                String word = queue.poll();
                for (String nextWord: getNextWords(word, dict)) {
                    if (hash.contains(nextWord)) {
                        continue;
                    }
                    if (nextWord.equals(end)) {
                        return length;
                    }
                    
                    hash.add(nextWord);
                    queue.offer(nextWord);
                }
            }
        }
        
        return 0;
    }

    // replace character of a string at given index to a given character
    // return a new string
    private String replace(String s, int index, char c) {
        char[] chars = s.toCharArray();
        chars[index] = c;
        return new String(chars);
    }
    
    // get connections with given word.
    // for example, given word = 'hot', dict = {'hot', 'hit', 'hog'}
    // it will return ['hit', 'hog']
    private ArrayList getNextWords(String word, Set dict) {
        ArrayList nextWords = new ArrayList();
        for (char c = 'a'; c <= 'z'; c++) {
            for (int i = 0; i < word.length(); i++) {
                if (c == word.charAt(i)) {
                    continue;
                }
                String nextWord = replace(word, i, c);
                if (dict.contains(nextWord)) {
                    nextWords.add(nextWord);
                }
            }
        }
        return nextWords;
    }
}