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