【题解】【字符串】【BFS】【Leetcode】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.

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：

字符串的花儿挺多，word ladder也是CC150 v5上的18.10题，乍一看无从下手，其实就是BFS求单源无权最短路径。

想要大set不超时，关键有几点：

1. 一边加新词进queue，一边从dict里remove掉

2. 直接寻找所有可能的OneEditWords判断是否在dict里面，每次都过一遍dict一个一个判断是否为OneEditWords会超时。我还有一个超时的方法，就是寻找所有可能的OneEditWords再建一个unordered_set跟dict求交集，后来发现algoritm里的set_intersection只支持两个有序集合很坑，需要先把unordered_set转化为vector排序，结果不言而喻。

3. 将unordered_map<string, int > path并到访问队列qwords里去会跑得更快，反正这里不要求输出路径，这个可以做下优化

4. 将getOneEditWords这个函数并到ladderLength里头会跑得更快，不过我觉得可读性会降低

int ladderLength(string start, string end, unordered_set<string> &dict) {
    unordered_set<string> ndict(dict);//最好不要修改输入参数
    queue<string> qwords;//BFS访问队列
    unordered_map<string, int > path;//记录到start的最短路径，其实可以并入qwords中
    
    qwords.push(start);
    path[start] = 1;
    ndict.erase(start);
    
    while(!qwords.empty()){
        start = qwords.front();
        qwords.pop();
        int len = path[start];
        for(string s :getOneEditWords(start)){//边局部构建map，边处理
            if(ndict.find(s) == ndict.end()) continue;//一个一个判断dict中元素是否为OneEditWords会超时
            if(s == end) return len+1;
            qwords.push(s);
            path[s] = len+1;
            ndict.erase(s);//如果不erase访问过的元素会超时
        }
    }
    return 0;    
} 

vector<string> getOneEditWords(string start){
    vector<string> words;
    for(unsigned int i = 0; i < start.size(); i++){
        string word(start);
        for(char ch = 'a'; ch <= 'z'; ch++){
            if(ch != start[i]){
                word[i] = ch;
                words.push_back(word);
            }
        }
    }
    return words;      
}