【LeetCode】127. Word Ladder

Bertha 。 2022-03-31 10:50 134阅读 0赞

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.

最短搜索路径，所以是广度优先搜索(BFS)。

有几个问题需要注意：

1、怎样判断是否为邻居节点？

按照定义，存在一个字母差异的单词为邻居，因此采用逐位替换字母并查找字典的方法寻找邻居。

（逐个比对字典里单词也可以做，但是在这题的情况下，时间复杂度会变高，容易TLE）

2、怎样记录路径长度？

对队列中的每个单词记录路径长度。从start进入队列记作1.

长度为i的字母的邻居，如果没有访问过，则路径长度为i+1

struct Node
    {
        string word;
        int len;
        
        Node(string w, int l): word(w), len(l) {}
    };
    
    class Solution {
    public:
        int ladderLength(string beginWord, string endWord, unordered_set<string>& wordDict) {
            queue<Node*> q;
            unordered_map<string, bool> m;
            Node* beginNode = new Node(beginWord, 1);
            q.push(beginNode);
            m[beginWord] = true;
            while(!q.empty())
            {
                Node* frontNode = q.front();
                q.pop();
                string frontWord = frontNode->word;
                //neighbor search
                for(int i = 0; i < frontWord.size(); i ++)
                {
                    for(char c = 'a'; c <= 'z'; c ++)
                    {
                        if(c == frontWord[i])
                            continue;
                            
                        string frontWordCp = frontWord;
                        frontWordCp[i] = c;
                        //end
                        if(frontWordCp == endWord)
                            return frontNode->len+1;
                        if(wordDict.find(frontWordCp) != wordDict.end() && m[frontWordCp] == false)
                        {
                            Node* neighborNode = new Node(frontWordCp, frontNode->len+1);
                            q.push(neighborNode);
                            m[frontWordCp] = true;
                        }
                    }
                }
            }
            return 0;
        }
    };

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[181128122607086.jpg]: /images/20220331/6210412ef56347a781f51f944fb86535.png