Word Ladder

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.

解答：BSF，第一次想用了一种从字典中找序列的思维，这个就和字典的大小有关系了，所以大数据不过。后来发现从单词出发开始做变化才是正解。

代码一：大数据不过

class Solution {
public:
    int ladderLength(string start, string end, unordered_set<string> &dict) {
        dict.erase(start);
        dict.erase(end);
        if(dict.size() == 0 && start != end) return 0;
        if(start == end) return 2;
       
        int n = dict.size() + 2;
        vector<int> vec(n,0);
        vector<vector<int>> matrix(n,vec);

        vector<string> vec1;
        vec1.push_back(start);
        vec1.insert(vec1.end(),dict.begin(),dict.end());
        vec1.push_back(end);
        
        for(int i = 0;i < n;i++)
        {
            for(int j = i+1;j < n;j++)
            {
                matrix[i][j] = calDis(vec1[i],vec1[j]);
		        matrix[j][i] = matrix[i][j];
            }
        }     
        queue<int> queue;
        int cur = 0;
        int cnt = 1;
        unordered_set<int> visited;
        queue.push(cur);
        queue.push(cnt);
        visited.insert(cur);
        while(!queue.empty())
        {
            cur = queue.front();
            queue.pop();
            cnt = queue.front();
            queue.pop();        
            for(int i = 0;i < n;i++)
            {
                if(visited.count(i) != 0 ||matrix[cur][i] == 0) continue;
                queue.push(i);
                queue.push(cnt+1);
		        visited.insert(cur);
            }   
            if(cur == n-1) {
                return cnt;
            }
        }
        return 0;  
    }
    
    int calDis(string s1,string s2)
    {
        int cnt = 0;
        for(int i = 0;i < s1.length();i++)
        {
            if(s1[i] != s2[i]) {
                if(++cnt > 1) return 0;
            }
        }
        return cnt;
    }
};

代码二：

class Solution {
public:
    int ladderLength(string start, string end, unordered_set<string> &dict) {

        if(dict.size() == 0 && start != end) return 0;
        if(start == end) return 2;
        
        queue<string> que;
        queue<int> que1;
        que.push(start);
        que1.push(0);
        dict.erase(start);
        
        while(!que.empty()){
            string cur = que.front();
            que.pop();           
            int cnt = que1.front();
            que1.pop();
            

            for(int i = 0;i < start.length();i++)
            {
                string tmp = cur;
                for(char j = 'a';j<='z';j++)
                {
                    tmp[i] = j;
                    if(tmp == end) return cnt+2;
                    if(dict.count(tmp) != 0){
                        que.push(tmp);
                        que1.push(cnt+1);
                        dict.erase(tmp);
                    }
                }
            } 
        }
      
        return 0;
    }
    
};

592 milli secs.

减少使用空间可以用变量代替queue.

class Solution {
public:
    int ladderLength(string start, string end, unordered_set<string> &dict) {

        if(dict.size() == 0 && start != end) return 0;
        if(start == end) return 2;
        
        queue<string> que;
        que.push(start);
        dict.erase(start);
        int lev1 = 1;
        int lev2 = 0;
        int cnt = 0;
        while(!que.empty()){
            string cur = que.front();
            que.pop();           
            --lev1;

            for(int i = 0;i < start.length();i++)
            {
                string tmp = cur;
                for(char j = 'a';j<='z';j++)
                {
                    tmp[i] = j;
                    if(tmp == end) return cnt+2;
                    if(dict.count(tmp) != 0){
                        que.push(tmp);
                        dict.erase(tmp);
						++lev2;
                    }
                }
            } 
			if (lev1 == 0) {
                ++cnt;
                lev1 = lev2;
                lev2 = 0;
            }
        }
      
        return 0;
    }
    
};