127. Word Ladder

Description

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.

Problem URL

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Solution

单词梯，给一个开始单词，和一个结束单词以及一个单词的列表，每次只变化一个字母，且必须变化为给定列表中的单词，问要几次才能变为endWord。

Using BFS to resolve this problem. We could maintain a set reached which is the word we have reached now. Converting wordlist to set wordDict to avoid TLE. We add beginWord to reached first. Then whild reached not contains endword, iteratation begins. Another hash set toAdd contains the words we could get from the wordDic after we change a character in reached word.

First for loop iterates words in reached set, for each word in reached, we convert it to a char array. Then for every characters in word, we change it to another character from a to z and determine whether it is in wordDict or not. If it is in wordDict, toAdd set adds this new word, and wordDict remove it. After all these procedures, if toAdd is empty, that means no path from begin word to end word, we could return 0. Or the iteration goes on, length ++ and reached = toAdd.

Code

class Solution {
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        Set<String> reached = new HashSet<String>();
        Set<String> wordDict = new HashSet<String>(wordList);
        reached.add(beginWord);
        wordList.add(endWord);
        int length = 1;
        while(!reached.contains(endWord)){
            Set<String> toAdd = new HashSet<String>();
            for (String word : reached){
                for (int i = 0; i < word.length(); i++){
                    char[] chars = word.toCharArray();
                    for (char ch = 'a'; ch <= 'z'; ch++){
                        chars[i] = ch;
                        String newWord = new String(chars);
                        if (wordDict.contains(newWord)){
                            toAdd.add(newWord);
                            wordDict.remove(newWord);
                        }
                    }
                }
            }
            
            length++;
            if (toAdd.size() == 0){
                return 0;
            }
            reached = toAdd;
        }
        return length;
    }
}

Time Complexity: unknow
Space Complexity: unknow

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