leetcode 127. Word Ladder

127. Word Ladder

这是一道广度优先遍历的题目

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:

1. Only one letter can be changed at a time.
2. Each transformed word must exist in the word list. Note that beginWord is not a transformed word.

For example,

Given:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log","cog"]

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.
• You may assume no duplicates in the word list.
• You may assume beginWord and endWord are non-empty and are not the same.

public class Solution {
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        Set<String> set=new HashSet<>();
        List<String> list=new ArrayList<>();
        set.add(beginWord);
        list.add(beginWord);
        if(beginWord.length()!=endWord.length())return 0;
        for(int i=0;i<wordList.size();i++){
            if(beginWord.length()==wordList.get(i).length()&&!set.contains(wordList.get(i))){
                set.add(wordList.get(i));
                list.add(wordList.get(i));
            }
        }
        int gcount=0;
        boolean mnark=false;
       Stack<String> stack=new Stack<>();
        Stack<String> mstack=new Stack<>();
        set=new HashSet<>();
        stack.push(beginWord);
        set.add(beginWord);
        list.remove(0);
        gcount++;
        while(!stack.empty()){
            String str=stack.pop();
            set.add(str);
            for(int i=0;i<list.size();i++){
                String temp=list.get(i);
                int count=0;
                for(int j=0;j<temp.length();j++){
                    if(temp.charAt(j)!=str.charAt(j))count++;
                }
                if(count==1){
                    mstack.push(temp);
                    list.remove(i);
                    i--;
                }
            }
            if(set.contains(endWord))return gcount;
            if(stack.empty()&&!mstack.empty()){
                stack=mstack;
                mstack=new Stack<>();
                gcount++;
            }
        }
        if(!mnark)return 0;
        return gcount;
    }
}