127.Word Ladder

題目：

從開始單詞到結束單詞，藉助字典是否可達，如果可達路徑長度是多少？

一個例子：

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

As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

思路：

1. 用一個集合來儲存現在那些單詞是可達的。BFS來尋找剩下的單詞。

2. 再說一個非常傻的想法，主要是爲了練習dijkstra算法的寫法。就是用這寫單詞計算出所有的鏈接關係，存在一張圖中，開始節點到結束節點的距離+1就是路徑長度。

那個很傻的程序的寫法：（會超時的！）

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.List;
import java.util.PriorityQueue;
import java.util.Set;


public class Solution {
	public boolean isconnected(String a, String b) {
		int cnt = 0;
		for(int i = 0; i < a.length(); i++) {
			if(a.charAt(i) != b.charAt(i))
				cnt++;
		}
		if(cnt == 1) return true;
		return false;
	}
	

	
    public int ladderLength(String beginWord, String endWord, Set<String> wordList) {
    	if(wordList.contains(beginWord))
    		wordList.remove(beginWord);
    	if(wordList.contains(endWord))
    		wordList.remove(endWord);
    	
    	int graphsize = wordList.size()+2;
    	int[][] graph = new int[graphsize][graphsize];
    	
    	List<Vertex> nodelist = new ArrayList<>();
    	nodelist.add(new Vertex(beginWord,0));
    	for(String wd: wordList) nodelist.add(new Vertex(wd,Integer.MAX_VALUE));
    	nodelist.add(new Vertex(endWord,Integer.MAX_VALUE));
    	
    	PriorityQueue<Vertex> Q =  new PriorityQueue<Vertex>(nodelist);
    	
    	/*step1 construct graph*/
    	for(int i = 0; i < graphsize; i++) {
    		for(int j = i+1; j < graphsize; j++)
    			if(isconnected(nodelist.get(i).content, nodelist.get(j).content)) {
    				graph[i][j] = 1;
    				graph[j][i] = 1;
    			}
    	}
    	
    	/*step2 dijkstra algorithm*/
    	while(!Q.isEmpty()) {
    		Vertex uVertex = Q.poll();

    		if(uVertex.dist != Integer.MAX_VALUE && uVertex.id == nodelist.size()-1) return uVertex.dist+1;


    		for(int i = 0; i < graphsize; i++) {
    			if(graph[uVertex.id][i] == 1) {
    				//relex two node

    				if(nodelist.get(i).dist > uVertex.dist+1) {
    					nodelist.get(i).dist = uVertex.dist+1;
    					nodelist.get(i).prev = uVertex;
    					Q.remove(nodelist.get(i));
    					Q.add(nodelist.get(i));
    				}
    				
    			}
    		}
    		
    	}
    	
    	return 0;

        
    }
    
    public static void main(String[] args) {
    	Set<String> wdlist = new HashSet<String>(Arrays.asList("dot","dog"));
    	System.out.println(new Solution().ladderLength("hot", "dog", wdlist));
    }
}

class Vertex implements Comparable<Vertex>{
	static int cnt = 0;
	int id;
	String content;
	Vertex prev = null;
	int dist = Integer.MAX_VALUE;
	
	public Vertex(String content, int dist) {
		this.content = content;
		this.dist = dist;
		this.id = cnt;
		cnt++;
	}

	@Override
	public int compareTo(Vertex o) {
		if(this.dist > o.dist) return 1;
		else if(this.dist < o.dist) return -1;
		else return 0;
	}
	
}

注意：

1. 初始化priorityqueue是直接用一個list初始化的，這些相當於只是把指針一個個移過去；

2. 當queue中元素的鍵值發生變化，他是不會自動在queue中移動位置的；你需要把他取出來在放進去一次。