[leetcode] Word Ladder

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:

1. Only one letter can be changed at a time
2. 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 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.

https://oj.leetcode.com/problems/word-ladder/

思路：BFS求最短路径。队列弹出字符串，每次变换一个字母，如果在字典中则插入队列，并从字典中删除。插入null来分层。

import java.util.HashSet;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Set;

public class Solution {
    public int ladderLength(String start, String end, Set<String> dict) {
        Queue<String> queue = new LinkedList<String>();
        dict.add(end);
        queue.add(start);
        queue.add(null);
        int len = start.length();
        int res = 1;

        while (!queue.isEmpty()) {
            String cur = queue.remove();

            if (cur != null && cur.equals(end)) {
                return res;
            }
            if (cur != null) {
                for (int i = 0; i < len; i++) {
                    char[] str = cur.toCharArray();
                    for (int j = 0; j < 26; j++) {
                        char ch = (char) ('a' + j);
                        if (str[i] != ch) {
                            str[i] = ch;
                            String changed = new String(str);
                            if (dict.contains(changed)) {
                                queue.add(changed);
                                dict.remove(changed);
                            }
                        }
                    }

                }
            } else if (!queue.isEmpty()) {
                res++;
                queue.add(null);
            }
        }
        return 0;

    }

    public static void main(String[] args) {
        Set<String> set = new HashSet<String>();
        String[] a = { "hot", "dot", "dog", "lot", "log" };
        for (String each : a)
            set.add(each);
        System.out.println(new Solution().ladderLength("hit", "log", set));
    }
}

 第二遍记录：

　　bfs的题目，注意null的用法，注意细节。

 

第三遍记录：

　　考虑各种边界情况，比如set是空的（end一开始会加进去，就不空了）

public class Solution {
    public int ladderLength(String start, String end, Set<String> dict) {
        dict.add(end);
        Queue<String> queue = new LinkedList<String>();
        queue.add(start);
        queue.add(null);
        int step = 1;
        int len =start.length();
        
        while(!queue.isEmpty()){
            String out = queue.remove();
            if(out==null){
                step++;
                if(!queue.isEmpty())
                    queue.add(null);        
            }else{
                if(out.equals(end)){
                    return step;
                }
                for(int i=0;i<len;i++){
                    char[] tmp = out.toCharArray();
                    for(int j=0;j<26;j++){
                        tmp[i]=(char)('a'+j);
                        String newStr= new String(tmp);
                        if(dict.contains(newStr)){
                            dict.remove(newStr);
                            queue.add(newStr);
                        }
                        
                    }          
                }    
            
            }
            
        
        }
        return 0;
        
    }

}