LeetCode | Word Ladder

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题目：

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence fromstart 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.

思路：

思路1：dijkstra算法，可以额外建立距离矩阵时间和空间开销太大。

思路2：BFS

代码：

思路1：

class Solution {public:    string start;    string end;    unordered_set<string> dict;        int ladderLength(string start, string end, unordered_set<string> &dict) {        this->start = start;        this->end = end;        this->dict = dict;                int **dist = new int*[dict.size()+2];        for(int k=0;k<dict.size()+2;k++)        {            dist[k]=new int[dict.size()+2];        }                int len=dict.size()+2;                for(int i=0;i<len;i++)        {            for(int j=i;j<len;j++)            {                if(i==j)                {                    dist[i][j]=INT_MAX;                }                else                {                    dist[i][j]=dist[j][i]=getDist(getString(i),getString(j));                }            }        }                int* minDist=new int[len];        bool* visited = new bool[len];        for(int i=0;i<len;i++)        {            minDist[i]=dist[0][i];            visited[i]=false;        }                while(minDist[dict.size()]==INT_MAX)        {            int min = INT_MAX;            int minD = INT_MAX;            for(int i=1;i<len;i++)            {                if(!visited[i]&&minDist[i]<minD)                {                    minD = minDist[i];                    min = i;                }            }            if(min==INT_MAX)            {                break;            }            else            {                visited[min]=true;                for(int i=1;i<len;i++)                {                    if(minDist[i]>minDist[min]+dist[min][i])                    {                        minDist[i]=minDist[min]+dist[min][i];                    }                }            }        }                return minDist[dict.size()]==INT_MAX?0:minDist[dict.size()];    }        int getDist(string s1, string s2)    {        int d=0;        for(int i;i<s1.size();i++)        {            if(s1[i]!=s2[i])            {                d++;            }        }        return d==1?1:INT_MAX;    }        string getString(int k)    {        if(k==0)        {            return start;        }        else if(k==dict.size()+1)        {            return end;        }        else        {            auto it = dict.begin();            while(k>1)            {                it++;                k--;            }            return *it;        }    }};

思路2：

class Solution {public:    int ladderLength(string start, string end, unordered_set<string> &dict) {        if(start==end)        {            return 1;        }                queue<string> nodes;        queue<int> depth;        nodes.push(start);        depth.push(1);                findItem(start,dict);        while(nodes.size()>0)        {            string cur = nodes.front();            nodes.pop();            int d = depth.front();            depth.pop();            for(int i=0;i<cur.size();i++)            {                for(int j=0;j<26;j++)                {                    if(cur[i]!='a'+j)                    {                        string tmp = cur;                        tmp[i] = 'a'+j;                        if(tmp==end)                        {                            return d+1;                        }                        if(findItem(tmp,dict))                        {                            nodes.push(tmp);                            depth.push(d+1);                        }                    }                }            }        }        return 0;    }        bool findItem(string s, unordered_set<string> &dict)    {        unordered_set<string>::const_iterator p = dict.find(s);        if(p!=dict.end())        {            dict.erase(p);            return true;        }        else        {            return false;        }            }};

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