leetcode--Word Ladder

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Problem Description:

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.

分析：根据题目的意思，我首先想到的是利用DFS每次查找到与当前string距离为1的string，用map来标记每次遍历过的string，下次不再比较，然后继续往下搜索，直到找到目标string，记录下距离len，然后与当前最短距离min比较，更新min，最后运行的结果是超时。后来在网上查了一下，才发现是用BFS，而且我的代码中没有充分利用set查找时间复杂度是O(1)的特性，每次都遍历map比较的时间复杂度达到O(mn)，其中m是字符串长度，n是set的元素个数。原始代码记录一下：

class Solution {public:        bool distance(string a,string b)    {        int dist=0;        for(int i=0;i<a.size();++i)        {            if(a[i]!=b[i])            dist++;        }        if(dist==1)            return true;        else            return false;    }        bool dfs(string start,string end,map<string,bool> &flag,int &res,int &min)    {        if(start==end)        {            if(res<min)                min=res;            return true;        }                for(map<string,bool>::iterator i=flag.begin();i!=flag.end();++i)        {            if((*i).second==true&&distance(start,(*i).first))            {                res++;                (*i).second=false;                dfs((*i).first,end,flag,res,min);                (*i).second=true;                res--;            }        }        return false;    }        int ladderLength(string start, string end, unordered_set<string> &dict) {        int res=0;        int len=dict.size();        if(len==0||start==end)            return 0;        int min=dict.size()+1;        map<string,bool> flag;        for(unordered_set<string>::iterator iter=dict.begin();iter!=dict.end();++iter)            flag.insert(make_pair(*iter,true));        bool tag=false;        tag=dfs(start,end,flag,res,min);        if(tag)            return min+1;        else            return 0;                    }};

Submission Result: Time Limit Exceeded

利用BFS遍历的代码如下：

class Solution {public:        int ladderLength(string start, string end, unordered_set<string> &dict) {        int res=0;        if(start==end||dict.size()==0)            return res;        unordered_set<string> midstr;//记录层次遍历的string，防止循环搜索        queue<string> strque;        strque.push(start);        int lev=1,count=0; //用于记录队列每层的元素个数        while(!strque.empty())        {            lev--;            string str=strque.front();            strque.pop();            if(str==end)            {                res++;                break;            }                        midstr.insert(str);            for(int i=0;i<str.size();++i)                for(char ch='a';ch<='z';++ch)                {                    string temp=str;                    if(str[i]==ch)                        continue;                    temp[i]=ch;                                            if(dict.count(temp)==1&&midstr.count(temp)==0)                    {                                                count++;                        strque.push(temp);                        midstr.insert(temp);                    }                }            if(lev==0)//当前层遍历完毕            {                res++;                lev=count;                count=0;            }        }         if(res==1)//如果是1说明不存在路径，按照题目的设置，没有路径为1的情况            return 0;        else            return res;    }};

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