LeetCode - 127. Word Ladder

题目描述

Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that:

Only one letter can be changed at a time.

Each transformed word must exist in the word list. Note that beginWord is not a transformed word.

Example

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

You may assume no duplicates in the word list.

You may assume beginWord and endWord are non-empty and are not the same.

原题链接

解题思路

经典的BFS题目。
想象一下，这个变换过程是一个树，
每一层是当前所有的变换结果，
下一层又是上一层的字符串的所有的变换结果。

例子：HIT
AIT, BIT, CIT, DIT.....
HAT, HBT, HCT, HDT.....
HIA, HIB, HIC, HID....
HIT 可以有这么多种变换方式，
而AIT, BIT本身也可以以相同的方式展开，
这就形成了一个相当大的树。

最直观的思路就是DFS，每次变成字典中的某个新单词，
同时从字典中删除这个单词然后不断递归。
但是大数据时候超时。因为要求的是最短的变换次数，
所以可以使用BFS，和DFS不一样不一次走到最深。
逐层遍历变换了一次、二次、三次、n次的所有单词。

思考一下DFS和BFS的区别，举个最简单的例子，111 -> 311。
DFS的话，111->112，之后需要DFS 112的所有变形。
同理111->113之后还要遍历113的所有变形。
而BFS的话，
111
112 113 121 131 211 311
只需要6次就能找到最终结果。

word ladder.jpg

Java代码实现

class Solution {
    public int ladderLength(String beginWord, String endWord, List wordList) {
        Set words = new HashSet(wordList);
        
        if (beginWord == null || endWord == null 
         || beginWord.length() == 0 || endWord.length() == 0
         || beginWord.length() != endWord.length() || !words.contains(endWord)) {
             return 0 ;
         }
        
        Queue  q = new LinkedList<>();
        q.offer(beginWord);
        
        Set visited = new HashSet<>();
        visited.add(beginWord);
        
        int level = 0;
        
        while (!q.isEmpty()) {
            int qSize = q.size();
            level++;
            
            for (int i = 0; i < qSize; i++) {
                String current = q.poll();
                for (int j = 0; j < current.length(); j++) {
                    char[] stringChar = current.toCharArray();
        
                    for (char c = 'a'; c <= 'z'; c++) {
                        stringChar[j] = c;

                        String temp = new String(stringChar);
                        
                        if (endWord.equals(temp)) {
                            return level + 1;
                        }
                        
                        if (words.contains(temp) && !visited.contains(temp)) {
                            visited.add(temp);
                            q.offer(temp);
                        }
                    }
                }
            }
        }
        return 0;
        
    }
}