LeetCode126:Word Ladder

题目:

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.
  • 解题思路:
  • 这题一开始没啥思路,谷歌一下,才知要用到BFS,既然要用到BFS,那当然形成一个抽象图。这里我们将每一个字符串当做图中一节点,如果两字符串只需通过变化一个字符即可相等,我们认为这两字符串相连。
  • 遍历图中节点时,我们通常会利用一个visit还标识是否访问过,这里我们将处理过的节点直接从dict中删除,以免重复处理。
  • 实现代码:
  • #include <iostream>
    #include <string>
    #include <queue>
    #include <unordered_set>
    using namespace std;
    
    /*
    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.
    */
    class Solution {
    public:
        int ladderLength(string start, string end, unordered_set<string> &dict) {
            if(start.empty() || end.empty() || dict.empty())
                return 0;
            queue<string> squ[2];//这里需要用到两个队列,因为是bfs,按层遍历,所以需要一层一层进行处理 
            squ[0].push(start);
            bool qid = false;
            int minLen = 1;
            while(!squ[qid].empty())
            {
                while(!squ[qid].empty())//处理同一层节点 
                {
                    string curstr = squ[qid].front();
                    squ[qid].pop();
                    for(int i = 0; i < curstr.size(); i++)
                    {
                        
                        for(char j = 'a'; j <= 'z'; j++)
                        {
                            if(j == curstr[i])
                                continue;
                            char t = curstr[i];
                            curstr[i] = j;
                            if(curstr == end)
                            {
                                return minLen+1;
                            }
                                
                            if(dict.count(curstr) > 0)
                            {
                                squ[!qid].push(curstr);
                                dict.erase(curstr);
                            }
                            curstr[i] = t;
                        }
                        
                    }
                                
                }
                qid = !qid;//表示将要处理的下一层 
                minLen++;
    
            }
            return 0;
            
        }
    };
    
    int main(void)
    {
        string start("hit");
        string end("cog");
        unordered_set<string> dict;
        dict.insert("hot");
        dict.insert("dot");
        dict.insert("dog");
        dict.insert("lot");
        dict.insert("log");
        Solution solution;
        int min = solution.ladderLength(start, end, dict);
        cout<<min<<endl;
        return 0;
    }
-----------------------我和我追猪的梦-----------------------------------------------------------------
  
作者:mickole
posted @ 2014-04-27 17:15  mickole  阅读(183)  评论(0编辑  收藏  举报