【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<||n!=end.size())
         {
             return ;
         }
         if(start==end)
         {
             return ;
         } 

         int level=;
         queue<string> q;
         q.push(start);
         //count用来记录每一个深度的元素的个数
         int count=;
         while()
         {
         start=q.front();
         q.pop();
         count--;
         for(int i=;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==)
         {
             level++;
             count=q.size();
         }
         }
         return ;
     }