【leetcode】Word Ladder

Word Ladder

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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.

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利用广度优先搜索，找到元素的深度即可

寻找相差一个字符的字符串时，考虑采用替换字符的方式寻找，遍历dict在dict很长时，会耗时

 

int ladderLength(string start, string end, unordered_set<string> &dict) {
        int n=start.size();
        if(n<1||n!=end.size())
        {
            return 0;
        }
        if(start==end)
        {
            return 1;
        } 
 
        int level=2;
        queue<string> q;
        q.push(start);   
        //count用来记录每一个深度的元素的个数    
        int count=1;       
        while(1)
        {          
        start=q.front();
        q.pop();
        count--;       
        for(int i=0;i<start.length();i++)
        {
            string ori=start;
//每次修改一个字符，看是否在字典中能找到
            for(char ch='a';ch<='z';ch++)
            {
                if(start[i]==ch)continue;
               
                start[i]=ch;
                if(start==end) return level;
      //如果能找到，则用queue记录下下一层深度的元素
                if(dict.find(start)!=dict.end())
                {
                    dict.erase(start);
                    q.push(start);
                }
                start=ori;
            }
        }
      
//没有下一层深度了，或者dict已经为空
        if(q.empty()||dict.empty())
        {
            break;
        }
       
        //count为0，说明该level的元素已经被遍历完了
        if(count==0)
        {
            level++;
            count=q.size();
        }    
        }
        return 0;
    }