Word Ladder 解答

Question

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 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.

Solution 1 -- BFS

 

class WordNode{
    String word;
    int numSteps;
    
    public WordNode(String word, int numSteps){
        this.word = word;
        this.numSteps = numSteps;
    }
}

public class Solution {
    public int ladderLength(String beginWord, String endWord, Set<String> wordList) {
        Queue<WordNode> queue = new LinkedList<WordNode>();
        queue.add(new WordNode(beginWord, 1));
        
        wordList.add(endWord);
        
        // 由于这里纪录了每个词的最小步数，所以不用两个list
        while (queue.size() > 0) {
            WordNode topNode = queue.remove();
            String current = topNode.word;
            
            if (current.equals(endWord))
                return topNode.numSteps;
            
            char[] arr = current.toCharArray();
            // 穷举法
            for (int i = 0; i < arr.length; i++) {
                for (char c = 'a'; c <= 'z'; c++) {
                    char tmp = arr[i];
                    if (arr[i] != c)
                        arr[i] = c;
                    
                    String newWord = new String(arr);
                    if (wordList.contains(newWord)) {
                        queue.add(new WordNode(newWord, topNode.numSteps + 1));
                        wordList.remove(newWord);
                    }
                    arr[i] = tmp;
                }
            }
        }
        return 0;
    }
}

 

Solution 2 -- BFS & Adjacency List

Basic idea is: 1. construct adjacency list 2. BFS

Constructing adjacency list uses O(n²) time, and BFS is O(n). Total time complexity is O(n²), when testing in leetcode, it reports "time limit exceeded".

public class Solution {
    public int ladderLength(String beginWord, String endWord, Set<String> wordList) {
        if (wordList == null)
            return -1;
        int step = 1;
        if (beginWord.equals(endWord))
            return 1;
        // Construct adjacency lists for words
        // Do not forget start word and end word
        wordList.add(beginWord);
        wordList.add(endWord);
        Map<String, List<String>> adjacencyList = new HashMap<String, List<String>>();
        Map<String, Boolean> visited = new HashMap<String, Boolean>();
        int length = wordList.size();
        String[] wordList2 = new String[length];
        int i = 0;
        for (String current : wordList) {
            wordList2[i] = current;
            i++;
        }
        // Initialization
        for (i = 0; i < length; i++) {
            adjacencyList.put(wordList2[i], new ArrayList<String>());
            visited.put(wordList2[i], false);
        }
        
        for (i = 0; i < length; i++) {
            String current = wordList2[i];
            // Construct adjacency list for each element
            for (int j = i + 1; j < length; j++) {
                String next = wordList2[j];
                if (isAdjacent(current, next)) {
                    List<String> list1 = adjacencyList.get(current);
                    list1.add(next);
                    adjacencyList.put(current, list1);
                    List<String> list2 = adjacencyList.get(next);
                    list2.add(current);
                    adjacencyList.put(next, list2);
                }
            }
        }
        
        // BFS
        List<String> current = new ArrayList<String>();
        List<String> next;
        current.add(beginWord);
        visited.put(beginWord, true);
        while (current.size() > 0) {
            step++;
            next = new ArrayList<String>();
            for (String currentString : current) {
                List<String> neighbors = adjacencyList.get(currentString);
                if (neighbors == null)
                    continue;
                for (String neighbor : neighbors) {
                    if (neighbor.equals(endWord))
                        return step;
                    if (!visited.get(neighbor)) {
                        next.add(neighbor);
                        visited.put(neighbor, true);
                    }
                }
            }
            current = next;
        }
        return -1;
        
    }
    
    private boolean isAdjacent(String current, String next) {
        if (current.length() != next.length())
            return false;
        int length = current.length();
        int diff = 0;
        for (int i = 0; i < length; i++) {
            char a = current.charAt(i);
            char b = next.charAt(i);
            if (a != b)
                diff++;
        }
        if (diff == 1)
            return true;
        return  false;
    }
}

 Python version

from collections import deque
from collections import defaultdict
    
class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
        if endWord not in wordList:
            return 0
        queue = deque()
        queue.append((beginWord, 1))
        visited = set([beginWord])
        adjacency_map = defaultdict(list)
        L = len(beginWord)
        # Pre-process
        for word in wordList:
            for i in range(L):
                adjacency_map[word[:i] + "*" + word[i+1:]].append(word)
        # BFS
        while queue:
            curr_word, curr_steps = queue.popleft()
            if curr_word == endWord:
                return curr_steps
            # List all possibilities
            for i in range(L):
                key = curr_word[:i] + "*" + curr_word[i+1:]
                if key in adjacency_map:
                    neighbors = adjacency_map[key]
                    for neighbor in neighbors:
                        if neighbor == endWord:
                            return curr_steps + 1
                        if neighbor not in visited:
                            queue.append((neighbor, curr_steps + 1))
                            visited.add(neighbor)
                    adjacency_map[key] = {}
        return 0