【leetcode】Word Ladder (hard) ★

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.

 

思路：

找两点之间的最小距离，感觉只能用图，构建图的邻接矩阵，然后folyd 果断的超时了.....

int ladderLength1(string start, string end, unordered_set<string> &dict) {

        if(start == end) return 1;

        int vexnum = dict.size() + 2;

        if(dict.find(start) != dict.end()) vexnum--;

        if(dict.find(end) != dict.end()) vexnum--;

        vector<string> s(vexnum); //记录graph中每行代表的单词

        vector<vector<int>> graph(vexnum, vector<int>(vexnum, vexnum + 1)); //邻接矩阵

        s[0] = start;

        s.back() = end;

        int i = 1;

        unordered_set<string>::iterator it;

        for(it = dict.begin(); it != dict.end(); it++)

        {

            if(*it != start && *it != end)

                s[i++] = (*it);

        }



        //记录有哪些单词可以通过一次变化相互转换

        for(i = 0; i < s.size(); i++)

        {

            string temp = s[i];

            for(int j = 0; j < start.size(); j++)

            {

                for(char c = 'a'; c <= 'z'; c++)

                {

                    temp[j] = c;

                    if(dict.find(temp) != dict.end())

                    {

                        int edge = find(s.begin(), s.end(), temp) - s.begin();

                        if(edge == i)

                        {

                            graph[i][edge] = 0;

                            graph[edge][i] = 0;

                        }

                        else

                        {

                            graph[i][edge] = 1;

                            graph[edge][i] = 1;

                        }

                    }

                }

            }

        }



        for(int k = 0; k < graph.size(); k++)

        {

            for(int i = 0; i < graph.size(); i++)

            {

                for(int j = 0; j < graph.size(); j++)

                {

                    if(graph[i][j] > graph[i][k] + graph[k][j])

                    {

                        graph[i][j] = graph[i][k] + graph[k][j];

                    }

                }

            }

        }



        return (graph[0][vexnum - 1] == vexnum + 1) ? 0 : graph[0][vexnum - 1] + 1;

    }

 

来看大神的BFS算法，用dis记录每个点到start的距离。队列里开始只有start, 然后每次遇到新转化成的单词就进队列，更新距离。判断能否转化时用字符长度和26个字母，而不是对字典遍历，因为字典中单词的数量可能远大于26个。

int ladderLength(string start, string end, unordered_set<string> &dict)

    {

        unordered_map<string, int> dis;

        queue<string> q;

        dis[start] = 1;

        q.push(start);

        while(!q.empty())

        {

            string word = q.front(); q.pop();

            for(int i = 0; i < start.size(); i++)

            {

                string temp = word;

                for(char c = 'a'; c <= 'z'; c++)

                {

                    temp[i] = c;

                    if(dict.count(temp) > 0 && dis.count(temp) == 0)

                    {

                        dis[temp] = dis[word] + 1;

                        q.push(temp);

                    }

                }

            }

        }

        if(dis.count(end) == 0) return 0;

        return dis[end];

    }