Word Ladder——LeetCode

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.

 

题目大意：给定两个等长的单词，各一个字典，每次只能变换一个字符得到中间词，此中间词必须在字典里存在，求从一个单词变换到另一个单词最短需要多少步。

解题思路：说实话这题一开始觉得不知道从哪儿下手，后来看了下tag是BFS，才有了思路，从字典里寻找与起始单词距离为1的所有单词，加入bfs队列，然后当队列非空，取出队列中的单词，查找在字典里的所有与之距离为1的单词，直到找到结束单词或者全部遍历完没有合法解，返回0；

Talk is cheap>>

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

        if (transInOne(start, end)) {

            return 2;

        }

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

        queue.add(start);

        int length = 1;

        while (!queue.isEmpty()) {

            length++;

            int size = queue.size();

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

                String toSearch = queue.poll();

                if (transInOne(toSearch, end)) {

                    return length;

                }

                for (Iterator<String> iterator = dict.iterator(); iterator.hasNext(); ) {

                    String next = iterator.next();

                    if (transInOne(toSearch, next)) {

                        queue.offer(next);

                        iterator.remove();

                    }

                }

            }

        }

        return 0;

    }



    private boolean transInOne(String start, String end) {

        int i = 0;

        int res = 0;

        while (i < start.length()) {

            if (start.charAt(i) != end.charAt(i)) {

                res++;

                if (res > 1)

                    return false;

            }

            i++;

        }

        return true;

    }