【leetcode】Word Ladder

Question ：

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.

For Example：

Graph Array =  [

h i t

h o t

d o t            

d o g           

l o t           

l o g

c o g

]

Anwser 1 ：     

class Solution {
public:
    int ladderLength(string start, string end, unordered_set<string> &dict) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(start == end) return 0;      // shortest length
        
        queue <pair<string, int>> Q;
        map<string, bool> hmap;
        vector<string> vec;
        
        Q.push(make_pair(start, 1));        // start word
        
        while(!Q.empty()){
            pair<string, int> node = Q.front();
            string word = node.first;
            int level = node.second;
            
            if(word == end){
                break;
            }
            Q.pop();
            vec.clear();
            getWords(vec, word, hmap, dict);
            
            int len = vec.size();
            for(int i = 0; i < len; ++i){
                Q.push(make_pair(vec[i], level + 1));
            }
        }
        
        if(Q.empty()) {
            return 0;
        } else {
            return Q.front().second;
        }
    }
    
    void getWords(vector<string> &vec, string start, map<string, bool> &hmap, const unordered_set<string> &dict){
        int n = start.size();
        for(int i = 0; i < n; ++i)
        {
            string temp(start);
            for(int j = 0; j < 26; ++j)
            {
                temp[i] = j + 'a';      // a - z
                if(temp != start && dict.find(temp) != dict.end() && hmap.find(temp) == hmap.end())
                {
                    hmap.insert(make_pair(temp, false));
                    vec.push_back(temp);
                }
            }
        }
    }
};

Anwser 2 ：       

class Solution {
public:
    int ladderLength(string start, string end, unordered_set<string> &dict) {        
        unordered_set<string> Set;
        queue<string> Q;
        int curLevel = 1;
        int curLevelNum = 1;
        int nextLevelNum = 0;
        
        Set.insert(start);
        Q.push(start);
        
        while (!Q.empty())
        {
            string cur = Q.front();
            Q.pop();
            
            --curLevelNum;
            for (unordered_set<string>::iterator it = dict.begin(); it != dict.end(); ++it)
            {
                string temp = *it;
                if (canChange(temp, cur) && Set.find(temp) == Set.end())
                {
                    Q.push(temp);
                    Set.insert(temp);
                    ++nextLevelNum;
                }
            }
            
            if (canChange(cur, end)) return curLevel + 1;
            
            if (curLevelNum == 0)
            {
                ++curLevel;
                curLevelNum = nextLevelNum;
                nextLevelNum = 0;                
            }
        }
        
        return 0;
    }
    
    bool canChange(string a, string b)
    {
        if (a.size() != b.size())
            return false;
        
        int count = 0;
        for (int i = 0; i < a.size(); ++i)
        {
            if (a[i] != b[i])            
            {
                if (count == 1) return false;   // more than 1 letters diff, return
                ++count;
            }
        }
        
        return count == 1;      // only one letter diff
    }
};

参考推荐：

unordered_set

set

queue

map

vector

pair