LeetCode 127 Word Ladder

A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that:

• Every adjacent pair of words differs by a single letter.
• Every si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList.
• sk == endWord
  Given two words, beginWord and endWord, and a dictionary wordList, return the number of words in the shortest transformation sequence from beginWord to endWord, or 0 if no such sequence exists.

Solution

注意到每次只能变一个字符，所以我们可以遍历每一个位置，将其设置为 \(\#\)，然后用 \(map\) 映射到 \(string\) 的 \(vector\) 里面。所以我们就可以利用 \(BFS\) 的方式，其中队列里面存储 \(pair(str, step)\)

点击查看代码

class Solution {
private:
    unordered_map<string, vector<string>> mp;
    queue<pair<string, int>> q;
    unordered_map<string,int> vis;
    
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        if(count(wordList.begin(), wordList.end(), endWord)==0)
            return 0;
        int n = wordList.size();
        wordList.push_back(beginWord);
        int m = wordList[0].size();
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                string tmp = wordList[i].substr(0,j)+'#'+wordList[i].substr(j+1,m-j-1);
                mp[tmp].push_back(wordList[i]);
            }
        }
        q.push({beginWord, 1});
        while(!q.empty()){
            auto f = q.front();q.pop();
            vis[f.first]=1;
            string cur = f.first;
            int cur_step = f.second;
            for(int i=0;i<m;i++){
                string tmp = cur.substr(0,i)+'#'+cur.substr(i+1,m-i-1);
                for(auto t:mp[tmp]){
                    if(!vis[t]){
                        if(t==endWord)return cur_step+1;
                        q.push({t, cur_step+1});
                    }
                }
            }
            
        }
        return 0;
    }
};