LeetCode 126 Word Ladder II (C++)

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Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) 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"]
Return
  [
    ["hit","hot","dot","dog","cog"],
    ["hit","hot","lot","log","cog"]
  ]
Note:
  • All words have the same length.
  • All words contain only lowercase alphabetic characters.
Solution C++


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class Solution {
    void generatePaths(vector<string>& path, vector<vector<string>>& ret, 
        const string& start, 
        const unordered_map<string, vector<string>>& parentmap) {
        if (path.back() == start) {
            vector<string> row = path;
            reverse(row.begin(), row.end());
            ret.push_back(row);
            return;
        }
        for (string s: parentmap.at(path.back())) {
            path.push_back(s);
            generatePaths(path, ret, start, parentmap);
            path.pop_back();
        }
    }
public:
    vector<vector<string>> findLadders(string start, string end, unordered_set<string> &dict) {
        vector<vector<string>> ret;
        if (start == end) {
            vector<string> path = {start, end};
            ret.push_back(path);
            return ret;
        }
        unordered_map<string, vector<string>> parentmap = {{start, vector<string>()}};
        unordered_set<string> current = {start};
        unordered_set<string> parent;
        dict.insert(end);
        bool matched = false;
        char letters[] = "abcdefghijklmnopqrstuvwxyz";
        while (!matched && !current.empty()) {
            parent = move(current);
            current.clear();
            for (string s: parent) {
                dict.erase(s);
            }
            while (!parent.empty()) {
                string parentstr = *(parent.begin());
                string curstr = parentstr;
                parent.erase(curstr);
                for (int i = 0; i < curstr.length(); ++i) {
                    char save = curstr[i];
                    for (char c: letters) {
                        if (save == c) continue;
                        curstr[i] = c;
                        if (dict.find(curstr) == dict.end()) continue;
                        current.insert(curstr);
                        parentmap[curstr].push_back(parentstr);
                        if (curstr == end) matched = true;
                    }
                    curstr[i] = save;
                }
            }
        }
        if (!matched) return ret;
        vector<string> path = {end};
        generatePaths(path, ret, start, parentmap);
        return ret;
    }
};

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