Leetcode050--字符串的过渡

一、原题

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:
Only one letter can be changed at a time
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 length5.

Note:
Return 0 if there is no such transformation sequence.
All words have the same length.
All words contain only lowercase alphabetic characters.

二、中文

给出一个开头字符串和一个结尾字符串还有一个字典，然后从字典中查询字符串，每次只能变换一个字母，问一共需要经历多少次才可以

三、举例

开始字符串 ="hit"
结尾字符串 ="cog"
字典 =["hot","dot","dog","lot","log"]

一个转移序列为 "hit" -> "hot" -> "dot" -> "dog" -> "cog",

返回的长度是 5 

四、思路

首先定义一个函数，判断两个字符串之间是否是相差一个字母，然后定义一个队列，从开始字符串开始进行字典的遍历比较，如果在字典中找到相差一个字母的字符串，那么就可以入队列了，然后将该字符串从字典中移除，最后得到的队列就是我们想要寻找的过程

五、程序

import java.util.*;
public class Solution {
public int ladderLength(String start, String end, HashSet<String> dict) {
int res = 1;
LinkedList<String> queue = new LinkedList();
queue.offer(start);
while(!queue.isEmpty()){
int size = queue.size();
while(size > 0){
String s = queue.poll();
size--;

if(isDiffone(s, end)){
return res + 1;
}
for(Iterator<String> it=dict.iterator();it.hasNext();){
String str = it.next();
if(isDiffone(str, s)){
queue.offer(str);
it.remove();
}
}
}
res++;
}
return 0;
}

public boolean isDiffone(String str1, String str2){
int count = 0;
for(int i = 0; i < str1.length(); i++){
if(str1.charAt(i) != str2.charAt(i)){
count++;
}
}
return (count == 1) ? true : false;
}
}