leetcode Word Ladder

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Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence frombeginWord to endWord, such that:

Only one letter can be changed at a time
Each intermediate word must exist in the word list

For example,

Given:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

思路1：一个map，一个队列，将每次转换的string加入到队列，并put到map中，距离+1，具体转换的方法网上其实有很多了，一般都是用将字符串的每一项分别用26个字母替换，并检查新的字符串是否在字典中，若是则添加到队列中，如果转换后得到endword，则返回距离，代码：

public int ladderLength(String beginWord, String endWord, Set<String> wordList) {    Map<String,Integer> map=new HashMap<>();    map.put(beginWord,1);    Queue<String> queue=new LinkedList<>();    queue.offer(beginWord);    while(!queue.isEmpty()){        String temp=queue.poll();        int dis=map.get(temp);        for(int i=0;i<temp.length();i++){            for(char j='a';j<='z';j++){                if(temp.charAt(i)==j) continue;                StringBuffer str=new StringBuffer(temp);                str.setCharAt(i,j);                String trans=str.toString();                if(trans.equals(endWord)) return dis+1;                if(wordList.contains(trans)&&!map.containsKey(trans)){                    queue.offer(trans);                    map.put(trans,dis+1);                }            }        }    }    return 0;}

这个方法实际run时有时会超时，有时会通过，非常神奇，说明运行时间处于允许范围的边缘。

思路2：用双集合法，这个方法是上面思路的一种改进，代码写的非常巧妙，极大缩减允许时间，直接上代码吧，只能膜拜，出处leetcode discuss：

public int ladderLength(String beginWord, String endWord, Set<String> wordList) {    Set<String> beginSet = new HashSet<String>(), endSet = new HashSet<String>();    int len = 1;    int strLen = beginWord.length();    HashSet<String> visited = new HashSet<String>();    beginSet.add(beginWord);    endSet.add(endWord);    while (!beginSet.isEmpty() && !endSet.isEmpty()) {        if (beginSet.size() > endSet.size()) {            Set<String> set = beginSet;            beginSet = endSet;            endSet = set;        }        Set<String> temp = new HashSet<String>();        for (String word : beginSet) {            char[] chs = word.toCharArray();            for (int i = 0; i < chs.length; i++) {                for (char c = 'a'; c <= 'z'; c++) {                    char old = chs[i];                    chs[i] = c;                    String target = String.valueOf(chs);                    if (endSet.contains(target)) {                        return len + 1;                    }                    if (!visited.contains(target) && wordList.contains(target)) {                        temp.add(target);                        visited.add(target);                    }                    chs[i] = old;                }            }        }        beginSet = temp;        len++;    }    return 0;}

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