Leetcode: 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.

BFS:

int ladderLength(string start, string end, unordered_set<string> &dict) {  
        // IMPORTANT: Please reset any member data you declared, as  
        // the same Solution instance will be reused for each test case.  
        if(start.size() != end.size()) return 0;  
        if(start.empty() || end.empty())return 0;  
          
        queue<string> path;  
        path.push(start);  
        int level = 1;  
        int count = 1;  
        dict.erase(start);  
        while(dict.size() > 0 && !path.empty())  
        {  
            string curword = path.front();  
            path.pop();count--;  
            for(int i = 0; i < curword.size(); i++)  
            {  
                string tmp = curword;  
                for(char j='a'; j<='z'; j++)  
                {  
                    if(tmp[i]==j)continue;  
                    tmp[i] = j;  
                    if(tmp==end)return level+1;  
                    if(dict.find(tmp) != dict.end()) path.push(tmp);  
                    dict.erase(tmp);  
                }  
            }  
            if(count==0)  
            {  
                count = path.size();  
                level++;  
            }  
        }  
        return 0;  
    }

补充：软件开发 , C++ ,