Word Ladder

22% Accepted

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:

Only one letter can be changed at a time
Each intermediate word must exist in the dictionary
Have you met this question in a real interview? Yes
Example
Given:
start = "hit"
end = "cog"
dict = ["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.

思路

• level order遍历， BFS的level order就是三重循环
• 解法里边用到的 string to char/ char to string 值得复习

String

public class Solution {
    public int ladderLength(String start, String end, Set<String> dict) {
        if (dict == null || dict.size() == 0) {
            return 0;
        }

        HashSet<String> hash = new HashSet<String>();
        Queue<String> queue = new LinkedList<String>();
        queue.offer(start);
        hash.add(start);

        int length = 1;
        /*
        典型的 BFS level order
         */
        while(!queue.isEmpty()) {
            length++;
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                String word = queue.poll();
                for (String nextWord: getNextWords(word, dict)) {
                    if (hash.contains(nextWord)) {
                        continue;
                    }
                    if (nextWord.equals(end)) {
                        return length;
                    }

                    hash.add(nextWord);
                    queue.offer(nextWord);
                }
            }
        }
        return 0;
    }

    // replace character of a string at given index to a given character
    // return a new string
    private String replace(String s, int index, char c) {
        char[] chars = s.toCharArray();
        chars[index] = c;
        return new String(chars);
    }

    // get connections with given word.
    // for example, given word = 'hot', dict = {'hot', 'hit', 'hog'}
    // it will return ['hit', 'hog']
    // 去循环每个字母，而不是单词，因为单词可能很多，而字母就这么些，我们查找用hashmap o(1)就能找到，省下了很多时间
    // 这get next words的整个过程时间复杂度 O(a*26*len) -> O(len)
    // 这样整个程序的时间复杂度 也就是 n*dict.size()*word.length
    private ArrayList<String> getNextWords(String word, Set<String> dict) {
        ArrayList<String> nextWords = new ArrayList<String>();
        for (char c = 'a'; c <= 'z'; c++) {
            for (int i = 0; i < word.length(); i++) {
                if (c == word.charAt(i)) {
                    continue;
                }
                String nextWord = replace(word, i, c);
                if (dict.contains(nextWord)) {
                    nextWords.add(nextWord);
                }
            }
        }
        return nextWords;
    }
}