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#127 Word Ladder

Description

A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> … -> sk such that:

Every adjacent pair of words differs by a single letter.
Every si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList.
sk == endWord
Given two words, beginWord and endWord, and a dictionary wordList, return the number of words in the shortest transformation sequence from beginWord to endWord, or 0 if no such sequence exists.

Examples

Example 1:

Input: beginWord = “hit”, endWord = “cog”, wordList = [“hot”,“dot”,“dog”,“lot”,“log”,“cog”]
Output: 5
Explanation: One shortest transformation sequence is “hit” -> “hot” -> “dot” -> “dog” -> cog", which is 5 words long.

Example 2:

Input: beginWord = “hit”, endWord = “cog”, wordList = [“hot”,“dot”,“dog”,“lot”,“log”]
Output: 0
Explanation: The endWord “cog” is not in wordList, therefore there is no valid transformation sequence.

Constraints:

1 <= beginWord.length <= 10
endWord.length == beginWord.length
1 <= wordList.length <= 5000
wordList[i].length == beginWord.length
beginWord, endWord, and wordList[i] consist of lowercase English letters.
beginWord != endWord
All the words in wordList are unique.

思路

这道题和 433 几乎是一模一样的,用第一种思路构建邻接矩阵,虽然能通过所有case,但由于5000*5000的邻接矩阵过大,最后会报TLE,但是通过word.length * 26个字母的遍历方式,就可以通过测试

另:使用HashSet进行contains判断,会比使用ArrayList进行contains判断要快很多很多很多,后者33/50样例就会TLE

另另:还有一种优化方式是从两头同时进行bi-search,这样可以缩小要检测的范围,代码我没有写,用的是acceptance里面的10ms解法

代码

邻接矩阵

class Solution {
    public boolean differOne(String a, String b){
        int count = 0;
        for (int i = 0; i < a.length(); i++)
            if (a.charAt(i) != b.charAt(i))
                count ++;
        return count == 1;
    }
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        if (beginWord.equals(endWord))
            return 1;
        
        int[][] matrix = new int[wordList.size()][wordList.size()];
        
        // handle special case
        boolean flag = false;
        for (String word: wordList){
            if (word.equals(endWord)){
                flag = true;
                break;
            }
        }
        if (!flag)
            return 0;
        
        for (int i = 0; i < wordList.size(); i++) {
            for (int j = i + 1; j < wordList.size(); j++) {
                if (differOne(wordList.get(i), wordList.get(j))){
                    matrix[i][j] = 1;
                    matrix[j][i] = 1;
                }
            }
        }
        
        List<Integer> toVisit = new ArrayList<>();
        int[] visited = new int[wordList.size()];
        int level = 2;
        
        for (int i = 0; i < wordList.size(); i++){
            if (differOne(beginWord, wordList.get(i))){
                if (wordList.get(i).equals(endWord))
                    return level;
                toVisit.add(i);
                visited[i] = 1;
            }
        }
        
        while (toVisit.size() > 0) {
            List<Integer> newVisit = new ArrayList<>();
            level ++;
            
            for (int num: toVisit){
                for (int i = 0; i < wordList.size(); i++){
                    if (matrix[i][num] == 1 && visited[i] == 0){
                        if (wordList.get(i).equals(endWord))
                            return level;
                        visited[i] = 1;
                        newVisit.add(i);
                    }
                }
            }
            
            toVisit = new ArrayList<>(newVisit);
        }
        
        return 0;
    }
}

word.length * 26

class Solution {
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        Set<String> wordDic = new HashSet<>(wordList);
        if (!wordDic.contains(endWord))
            return 0;
        
        String[] alp = new String[]{"a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s","t", "u", "v", "w", "x", "y", "z"};
        
        List<String> toVisit = new ArrayList<>();
        Set<String> visited = new HashSet<>();
        
        toVisit.add(beginWord);
        
        int level = 1;
        
        while (toVisit.size() > 0){
            level ++;
            List<String> newToVisit = new ArrayList<>();
            
            for (String str: toVisit) {
                for (int i = 0; i < str.length(); i++) {
                    for (String a: alp) {
                        StringBuilder sb = new StringBuilder(str);
                        sb.replace(i, i + 1, a);
                        String after = sb.toString();
                        
                        if (wordDic.contains(after) && !visited.contains(after)){
                            if (after.equals(endWord))
                                return level;
                            newToVisit.add(after);
                            visited.add(after);
                        }
                    }
                }
            }
            
            toVisit = new ArrayList<>(newToVisit);
        }
        
        return 0;
    }
}

bi-Search

class Solution {
    
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        // 第 1 步:先将 wordList 放到哈希表里,便于判断某个单词是否在 wordList 里
        Set<String> wordSet = new HashSet<>(wordList);
        if (wordSet.size() == 0 || !wordSet.contains(endWord)) {
            return 0;
        }

        // 第 2 步:已经访问过的 word 添加到 visited 哈希表里
        Set<String> visited = new HashSet<>();
        
        // 分别用左边和右边扩散的哈希表代替单向 BFS 里的队列,它们在双向 BFS 的过程中交替使用
        Set<String> beginVisited = new HashSet<>();
        beginVisited.add(beginWord);
        Set<String> endVisited = new HashSet<>();
        endVisited.add(endWord);

        // 第 3 步:执行双向 BFS,左右交替扩散的步数之和为所求
        int step = 1;
        while (!beginVisited.isEmpty() && !endVisited.isEmpty()) {
            // 优先选择小的哈希表进行扩散,考虑到的情况更少
            if (beginVisited.size() > endVisited.size()) {
                Set<String> temp = beginVisited;
                beginVisited = endVisited;
                endVisited = temp;
            }

            // 逻辑到这里,保证 beginVisited 是相对较小的集合,nextLevelVisited 在扩散完成以后,会成为新的 beginVisited
            Set<String> nextLevelVisited = new HashSet<>();
            for (String word : beginVisited) {
                if (changeWordEveryOneLetter(word, endVisited, visited, wordSet, nextLevelVisited)) {
                    return step + 1;
                }
            }

            // 原来的 beginVisited 废弃,从 nextLevelVisited 开始新的双向 BFS
            beginVisited = nextLevelVisited;
            step++;
        }
        return 0;
    }


    /**
     * 尝试对 word 修改每一个字符,看看是不是能落在 endVisited 中,扩展得到的新的 word 添加到 nextLevelVisited 里
     *
     * @param word
     * @param endVisited
     * @param visited
     * @param wordSet
     * @param nextLevelVisited
     * @return
     */
    private boolean changeWordEveryOneLetter(String word, Set<String> endVisited,
                                             Set<String> visited,
                                             Set<String> wordSet,
                                             Set<String> nextLevelVisited) {
        char[] charArray = word.toCharArray();
        for (int i = 0; i < word.length(); i++) {
            char originChar = charArray[i];
            for (char c = 'a'; c <= 'z'; c++) {
                // if (originChar == c) {
                //     continue;
                // }
                charArray[i] = c;
                String nextWord = String.valueOf(charArray);
                if (wordSet.contains(nextWord)) {
                    if (endVisited.contains(nextWord)) {
                        return true;
                    }
                    if (!visited.contains(nextWord)) {
                        nextLevelVisited.add(nextWord);
                        visited.add(nextWord);
                    }
                }
            }
            // 恢复,下次再用
            charArray[i] = originChar;
        }
        return false;
    }
}

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