#leetcode#Word Ladder

Given two words (beginWord and endWord), and a dictionary, 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 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.

学习了Code Ganker大神的代码。。

BFS， lastNum标示上一层的节点数， curNum标示当前层的节点数， level标示到了第几层，

代码中有一个地方注意， 最后return 0， 因为有解的情况在循环中已经处理了，，

还有就是String的比较用String.equals（）

public class Solution {
    public int ladderLength(String beginWord, String endWord, Set<String> wordDict) {
        if(beginWord == null || beginWord.length() == 0 || endWord == null || endWord.length() == 0 || wordDict == null || wordDict.size() == 0){
            throw new IllegalArgumentException("invalid input");
        }
        int lastNum = 1;
        int curNum = 0;
        int level  = 1;
        LinkedList<String> queue = new LinkedList<String>();
        Set<String> visited = new HashSet<String>();
        queue.offer(beginWord);
        visited.add(beginWord);
        while(!queue.isEmpty()){
            String curWord = queue.poll();
            lastNum--;
            for(int i = 0; i < curWord.length(); i++){
                char[] arr = curWord.toCharArray();
                for(char c = 'a'; c <='z'; c++){
                    arr[i] = c;
                    String tmp = new String(arr);
                    if(tmp.equals(endWord)){
                        level++;
                        return level;
                    }
                    if(wordDict.contains(tmp) && !visited.contains(tmp)){
                        curNum++;
                        visited.add(tmp);
                        queue.offer(tmp);
                    }
                }
            }
            if(lastNum == 0){
                lastNum = curNum;
                curNum = 0;
                level++;
            }
        }
        
        // return level;
        return 0; // valid solution not found.. return 0;
    }
}