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
For 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.
Analysis: As for a shortest path problem like this one, BFS is always the best solution. Note that we need to maintain a visited list to store the words that have been visited previously.
public class Solution {
public int ladderLength(String start, String end, HashSet<String> dict) {
HashMap<String, Boolean> visited = new HashMap<String, Boolean>();
Queue<String> frontier = new LinkedList<String>();
visited.put(start, true);
frontier.add(start);
int transformation = 1;
while(!frontier.isEmpty()) {
Queue<String> next = new LinkedList<String>();
while(!frontier.isEmpty()) {
if(frontier.peek().equals(end)) return transformation;
char[] current = frontier.poll().toCharArray();
for(int i=0; i<current.length; i++) {
char holder = current[i];
for(char j='a'; j<='z'; j++) {
if(j != current[i]) {
current[i] = j;
String tem = new String(current);
if(dict.contains(tem) && !visited.containsKey(tem)) {
next.add(tem);
visited.put(tem, true);
}
}
}
current[i] = holder;
}
}
frontier = next;
transformation++;
}
return 0;
}
}