【LeetCode】127. Word Ladder-蒲公英云

【LeetCode】127. Word Ladder

Bertha 。 2022-03-31 10:50 267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:

Only one letter can be changed at a time
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;
    }
};