Leetcode 127. Word Ladder(java)

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 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:

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:

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.

解法一：BFS，不解释

class Solution {
    //uni-directional BFS
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        int len = 0;
        Set<String> set = new HashSet<>();
        for (int i = 0; i < wordList.size(); i++) {
            set.add(wordList.get(i));
        }
        if (!set.contains(endWord)) return 0;
        Queue<String> q = new LinkedList<>();
        q.offer(beginWord);
        while (!q.isEmpty()) {
            len++;
            int size = q.size(); 
            while (size-- > 0) {
                StringBuilder cur = new StringBuilder(q.poll());
                for (int i = 0; i < cur.length(); i++) {
                    char c = cur.charAt(i);
                    for (int j = 0; j < 26; j++) {
                        if (c == (char)('a' + j)) continue;
                        cur.setCharAt(i, (char)('a' + j));
                        String str = cur.toString();
                        if (set.contains(str)) {
                            if (str.equals(endWord)) {
                                return len + 1;
                            }
                            q.offer(str);
                            set.remove(str);
                        }
                        cur.setCharAt(i, c);
                    }
                }
            }
        }
        return 0;
    }
}

解法二：Bidirectional BFS

这个写法还有点麻烦了，实际上要理解的是，每个方向都只在乎一个前进方向的最前一排node和另一个方向的最前一排node，因此只需要q1, q2和前进的时候新建的q，每次前进一排后交换q和q1, 然后每次前进之前判断q1和q2的大小（如果q1>q2，交换q1和q2），保证每次都用前排数量最小的排前进即可。

class Solution {
    //Bi-directional BFS
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        HashSet<String> set = new HashSet<>();
        for (int i = 0; i < wordList.size(); i++) {
            set.add(wordList.get(i));
        }
        if (!set.contains(endWord)) return 0;
        HashSet<String> s1 = new HashSet<>();
        HashSet<String> s2 = new HashSet<>();
        List<String> next1 = new ArrayList<>();
        List<String> next2 = new ArrayList<>();
        s2.add(beginWord);
        s1.add(endWord);
        next1.add(endWord);
        next2.add(beginWord);
        int result = 0;
        while (!next2.isEmpty()) {
            result++;
            //swap s1 and s2
            HashSet<String> temp = s1;
            s1 = s2;
            s2 = temp;
            List<String> cur = next1;
            next1 = next2;
            next2 = cur;
            
            int size = next1.size();
            while (--size >= 0) {
                StringBuilder s = new StringBuilder(next1.get(size));
                for (int i = 0; i < s.length(); i++) {
                    char c = s.charAt(i);
                    for (int j = 0; j < 26; j++) {
                        if (c == (char)(j+'a')) continue;
                        s.setCharAt(i, (char)(j+'a'));
                        String str = s.toString();
                        if (s2.contains(str)) {
                            return result + 1;
                        }
                        if (set.contains(str)) {
                            next1.add(str);
                            s1.add(str);
                            set.remove(str);
                        }
                    }
                    s.setCharAt(i, c);
                }
                next1.remove(next1.get(size));
            }
        }
        return 0;
    }
}