leetcode——Word Ladder II

class Solution {
    // 用广度搜索，建立由start开始的搜索树，子节点为父节点只改变一个字母在字典里能查到的所有的单词
    // 每个子节点能记录其父节点及其所在最小层数，保证最短路径
    // 建立完树之后，遍历树，若结点能转换为end，则将其到start的路径记录，并加入sequences
private:
    vector<vector<string> > sequences; // 记录结果
public:
    vector<vector<string>> findLadders(string start, string end, unordered_set<string> &dict) {
        unordered_map<string, pair<queue<string>, int> > father; // 记录父节点及层数
        unordered_set<string> visited; // 记录已访问过
        vector<string> sequence;
        queue<string> que;
        int shortest(INT_MAX); // 到end最短的序列
        if (start.size() != end.size() || start.size() < 1) return sequences;
        int size = start.size();
        if (size == 1) {
            sequence.push_back(start);
            sequence.push_back(end);
            sequences.push_back(sequence);
            return sequences;
        }
        // 访问根节点入队，记录父节点极其层数
        que.push(start); 
        queue<string> fq; fq.push("");
        father[start] = make_pair(fq, 1);
        while (!que.empty()){
            // 出队
            string word = que.front();
            que.pop();
            // 改变字母
            for (int i = 0; i < size; i++){
                for (char c = ‘a‘; c <= ‘z‘; c++){
                    if (c == word[i]) continue;
                    string newWord = word;
                    newWord    = newWord.replace(i, 1, 1, c);
                    // 如果是end，则若是第一次出现则是最短路径
                    // 因为是一层一层地搜索，所以第一次搜索到end则是最短的
                    // 若是最短路径，则记录路径，加入sequences
                    if (newWord == end) {
                        if (shortest == INT_MAX) shortest = father[word].second + 1;
                        int curLayer = father[word].second + 1;
                        if (curLayer = shortest) addSequence(father, start, end, word);
                    }
                    // 如果在字典里且没访问过
                    if (dict.find(newWord) != dict.end() && visited.find(newWord) == visited.end())
                    {
                        // 入队，记录其所在层及其父节点
                        int layer = father[word].second;
                        // 若newWord没有父节点,则创建
                        if (father.find(newWord) == father.end()){
                            que.push(newWord);
                            queue<string> fq;
                            fq.push(word);
                            father[newWord] = make_pair(fq, layer + 1);
                        }//若newWord已有父节点，且newWord的层数比现在word的层数多1，则增加其父节点
                        else    if (layer <= (father[newWord].second - 1)){
                            que.push(newWord);
                            auto fq = father[newWord].first;
                            fq.push(word);
                            father[newWord] = make_pair(fq,layer + 1);
                        }// 若newWord有父节点，且当前层数大于其最小层数，则标记其为访问过
                        else{
                            visited.insert(newWord);
                        }
                        
                    }
                }
            }
        }
        return sequences;
    }
    inline void addSequence(unordered_map<string, pair<queue<string>, int> > &father, string start, string end, string word){
        vector<string> sequence;
        if(word != end) sequence.push_back(word);
        sequence.push_back(end);
        auto wordfather = father[word].first;
        auto fatherWord = wordfather.front();
        auto n = father[word].second;
        if (wordfather.size() > 1) {
            wordfather.pop();
            father[word] = make_pair(wordfather, n);
        }
        word = fatherWord;
        while (word != start && word != ""){
            sequence.insert(sequence.begin(), word);
            wordfather = father[word].first;
            auto fatherWord = wordfather.front();
            auto n = father[word].second;
            if (wordfather.size() > 1) {
                wordfather.pop();
                father[word] = make_pair(wordfather, n);
            }
            word = fatherWord;
        }
        if (sequence[0] != start) sequence.insert(sequence.begin(), start);
        sequences.push_back(sequence);
    }
};