Word Ladder

Word Ladder

问题:

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

思路：

　　层次遍历

我的代码：

public class Solution {

    public int ladderLength(String start, String end, Set<String> dict) {

        if(isValid(start, end)) return 2;

        Set<String> rst = new HashSet<String>();

        for(String str : dict)

        {

            if(isValid(end,str))

            {

                rst.add(str);

            }

        }

        Queue<String> queue = new LinkedList<String>();

        queue.add(start);

        int level = 2;

        while(!queue.isEmpty())

        {

            int size = queue.size();

            for(int i = 0; i < size; i++)

            {

                String tmp = queue.poll();

                for(char c = 'a'; c <= 'z'; c++)

                {

                    for(int j = 0; j < tmp.length(); j++)

                    {

                        if(tmp.charAt(j) == c)  continue;

                        String str = replace(tmp, j, c);

                        if(rst.contains(str))  

                        {

                            return level + 1;

                        }

                        if(dict.contains(str))

                        {

                            queue.offer(str);

                            dict.remove(str);

                        }

                        

                    }

                }

            }

            level++;

        }

        return 0;

    }

    private String replace(String s, int index, char c)

    {

        char[] chars = s.toCharArray();

        chars[index] = c;

        return new String(chars);

    }

    public boolean isValid(String one, String two)

    {

        int count = 0;

        for(int i = 0; i < one.length(); i++)

        {

            if(one.charAt(i) != two.charAt(i))

                count++;

        }

        return count == 1 ? true : false;

    }

}

View Code

• 层次遍历的应用之一，学习之处在于如何改变一个字符串的一位。