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;

}