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leetcode 127. word ladder

admin 11月 13, 2020 0

127. Word Ladder

Difficulty:: Medium

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:

  1. Only one letter can be changed at a time.
  2. Each transformed word must exist in the word list. Note that beginWord is not a transformed word.

Note:

  • Return 0 if there is no such transformation sequence.
  • All words have the same length.
  • All words contain only lowercase alphabetic characters.
  • You may assume no duplicates in the word list.
  • You may assume beginWord and endWord are non-empty and are not the same.

Example 1:

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

Output: 5

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

Example 2:

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

Output:  0

Explanation:  The endWord "cog" is not in wordList, therefore no possible  transformation.

Solution

Language: Java

通过修改word中的字母生成 78%

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class  {
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        if (beginWord == null || endWord == null || wordList == null || wordList.size() == 0) {
            return 0;
        }
        Set<String> wordSet = new HashSet<>(wordList);
        if (!wordSet.contains(endWord)) {
            return 0;
        }
        Queue<String> q = new LinkedList<>();
        q.offer(beginWord);
        int length = 1;
        while (!q.isEmpty()) {
            int size = q.size();
            for (int i = 0; i < size; i++) {
                String curWord = q.poll();
                for (String next : nextWords(curWord, wordSet)) {
                    if (next.equals(endWord)) {
                        return length + 1;
                    }
                    q.offer(next);
                }
            }
            length++;
        }
        return 0;
        
    }
    
    private List<String> nextWords(String word, Set<String> wordSet) {
        List<String> nextWords = new ArrayList<>();
        char[] chars = word.toCharArray();
        for (int i = 0; i < chars.length; i++) {
            char origin = chars[i];
            for (char c = 'a'; c <= 'z'; c++) {
                if (c == origin) {
                    continue;
                }
                chars[i] = c;
                String newStr = new String(chars);
                if (wordSet.contains(newStr)) {
                    nextWords.add(newStr);
                    wordSet.remove(newStr);
                }
            }
            chars[i] = origin;
        }
        return nextWords;
    }
}
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