LeetCode 127. Word Ladder（单词梯子）

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原题网址：https://leetcode.com/problems/word-ladder/

Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord 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.

Note:

Return 0 if there is no such transformation sequence.
All words have the same length.
All words contain only lowercase alphabetic characters.

方法：广度优先搜索。

public class Solution {    public int ladderLength(String beginWord, String endWord, Set<String> wordList) {        if (beginWord.equals(endWord)) return 1;        Map<String, String[]> masks = new HashMap<>();        Map<String, List<String>> ladders = new HashMap<>();        wordList.add(beginWord);        wordList.add(endWord);        Iterator<String> it = wordList.iterator();        while (it.hasNext()) {            String word = it.next();            char[] wa = word.toCharArray();            String[] mask = new String[wa.length];            for(int i=0; i<mask.length; i++) {                char ch = wa[i];                wa[i] = '*';                mask[i] = new String(wa);                wa[i] = ch;                List<String> ladder = ladders.get(mask[i]);                if (ladder == null) {                    ladder = new ArrayList<>();                    ladders.put(mask[i], ladder);                }                ladder.add(word);            }            masks.put(word, mask);        }        List<String> currents = new ArrayList<>();        currents.add(beginWord);        Set<String> visited = new HashSet<>();        visited.add(beginWord);        // wordList.remove(beginWord);        int dist = 0;        while (!currents.isEmpty()) {            dist ++;            if (currents.contains(endWord)) return dist;            List<String> nexts = new ArrayList<>();            for(int i=0; i<currents.size(); i++) {                String current = currents.get(i);                String[] mask = masks.get(current);                for(int j=0; j<mask.length; j++) {                    List<String> ladder = ladders.get(mask[j]);                    for(int k=0; k<ladder.size(); k++) {                        String next = ladder.get(k);                        if (visited.contains(next)) continue;                        // if (!wordList.contains(next)) continue;                        nexts.add(next);                        visited.add(next);                        // wordList.remove(next);                    }                }            }            currents = nexts;        }        return 0;    }}

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