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;    }}