leetCode练习（127）

题目：Word Ladder

难度：medium

问题描述：

Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence frombeginWord to endWord, such that:

1. Only one letter can be changed at a time
2. Each intermediate word must exist in the word list

For example,

Given:
beginWord = "hit"
endWord = "cog"
wordList = ["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.

解题思路：

可以看做图的最短路径问题，使用BFS即可。在两个String判断是否相邻时要注意，对wordlist，不要用枚举，因为wordlist的量可能非常非常大，会超时。要对另一个String进行修改后对wordList进行contains判定才能不超时、

代码如下：

public class Solution {    public int ladderLength(String beginWord, String endWord, Set<String> wordList) {        wordList.add(endWord);Queue<String> queue=new LinkedList<String>();queue.add(beginWord);int level=0;while(!queue.isEmpty()){int len=queue.size();for(int i=0;i<len;i++){String temp=queue.poll();if(temp.equals(endWord)){return level+1;}for(int j=0;j<temp.length();j++){    char[] cs=temp.toCharArray();for(char c='a';c<'z';c++){cs[j]=c;String check=new String(cs);if(!check.equals(temp)&&wordList.contains(check)){queue.add(check);wordList.remove(check);}}}}level++;}return 0;    }}

 

另外，我也造轮子解了一次题~并且犯了上面说的错误，妥妥的超时、

轮子代码如下：

public class node{String Value;HashSet<node> nodelist;public node(String v){nodelist=new HashSet<>();Value=v;}public void add(node x){nodelist.add(x);}}public class Graph{public ArrayList<node> graphSet;public Graph(HashSet<node> nodset){graphSet=new ArrayList<>(nodset);}     public int route(String s1,String s2){    node n1,n2;    n1=new node("");    n2=new node("");    for(node w:graphSet){    if(w.Value.equals(s1)){    n1=w;    }    if(w.Value.equals(s2)){    n2=w;    }    }    Queue<node> queue=new LinkedList<>();    queue.add(n1);    int count=0;    while(queue.size()>0){    int len=queue.size();    for(int i=0;i<len;i++){    node temp=queue.poll();    if(isnext(temp.Value,n2.Value)){    return count+2;    }    for(node w:graphSet){    if(isnext(temp.Value,w.Value)){    queue.add(w);    }    }    }    count++;    }    return 0;    }}public Graph getGraph(String beginWord, String endWord, Set<String> wordList){HashSet<node> nodset=new HashSet<node>();for(String w:wordList){node temp=new node(w);nodset.add(temp);}nodset.add(new node(beginWord));nodset.add(new node(endWord));for(node w:nodset){for(node x:nodset){if(isnext(w.Value,x.Value)){if(!w.nodelist.contains(x)){w.nodelist.add(x);}if(!x.nodelist.contains(w)){x.nodelist.add(w);}}}}return new Graph(nodset);}public static boolean isnext(String s1,String s2){if(s1.length()!=s2.length())return false;int count=0;for(int i=0;i<s1.length();i++){if(s1.charAt(i)!=s2.charAt(i)){count++;}}return count==1;}