LeetCode(126) Word Ladder II

题目如下：

Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that:

Only one letter can be changed at a time
Each intermediate word must exist in the dictionary
For example,

Given:
start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]
Return

分析如下：
这道题目真心很难通过。

1 考虑保存father信息来记录每个节点的前驱。

2 可是，最优路径可能在某个节点重叠。这样一个节点就不止一个father。

3 如果考虑用一个list或者vector保存当前节点的前驱，那么可能会Memory Limit Exceed

4 haoel大牛的做法是BFS + DFS.

// For example:////     start = "hit"//     end = "cog"//     dict = ["hot","dot","dog","lot","log","dit","hig", "dig"]////                      +-----+//        +-------------+ hit +--------------+//        |             +--+--+              |//        |                |                 |//     +--v--+          +--v--+           +--v--+//     | dit |    +-----+ hot +---+       | hig |//     +--+--+    |     +-----+   |       +--+--+//        |       |               |          |//        |    +--v--+         +--v--+    +--v--+//        +----> dot |         | lot |    | dig |//             +--+--+         +--+--+    +--+--+//                |               |          |//             +--v--+         +--v--+       |//        +----> dog |         | log |       |//        |    +--+--+         +--+--+       |//        |       |               |          |//        |       |    +--v--+    |          |//        |       +--->| cog |<-- +          |//        |            +-----+               |//        |                                  |//        |                                  |//        +----------------------------------+

haoel大牛的代码：

// 654msclass Solution {public:    map< string, unordered_set<string> >& buildTree (            string& start,            string& end,            unordered_set<string> &dict) {                static map< string, unordered_set<string> > parents;        parents.clear();                unordered_set<string> level[3];        unordered_set<string> *previousLevel = &level[0];        unordered_set<string> *currentLevel = &level[1];        unordered_set<string> *newLevel = &level[2];        unordered_set<string> *p =NULL;        currentLevel->insert(start);                bool reachEnd = false;                while( !reachEnd ) {            newLevel->clear();            for(auto it=currentLevel->begin(); it!=currentLevel->end(); it++) {                for (int i=0; i<it->size(); i++) {                    string newWord = *it;                    for(char c='a'; c<='z'; c++){                        newWord[i] = c;                        if (newWord == end){                            reachEnd = true;                            parents[*it].insert(end);                            continue;                        }                        if ( dict.count(newWord)==0 || currentLevel->count(newWord)>0 || previousLevel->count(newWord)>0 ) {                            continue;                        }                        newLevel->insert(newWord);                        //parents[newWord].insert(*it);                        parents[*it].insert(newWord);                    }                }            }            if (newLevel->empty()) break;                        p = previousLevel;            previousLevel = currentLevel;            currentLevel = newLevel;            newLevel = p;        }                if ( !reachEnd ) {            parents.clear();        }        return parents;    }            void generatePath(string start, string end,            map< string, unordered_set<string> > &parents,//            vector<string> path,            vector<string> &path, //减少内存使用，发挥DFS优势。            vector< vector<string> > &paths) {                if (parents.find(start) == parents.end()){            if (start == end){                paths.push_back(path);            }            return;        }                for(auto it=parents[start].begin(); it!=parents[start].end(); it++){            path.push_back(*it);            generatePath(*it, end, parents, path, paths);            path.pop_back();        }            }            vector< vector<string> >  findLadders(            string start,            string end,            unordered_set<string> &dict) {                vector< vector<string> > ladders;        vector<string> ladder;        ladder.push_back(start);        if (start == end){            ladder.push_back(end);            ladders.push_back(ladder);            return ladders;        }                map< string, unordered_set<string> >& parents = buildTree(start, end, dict);                if  (parents.size()<=0) {            return ladders;        }                generatePath(start, end, parents, ladder, ladders);                return ladders;    }    };

参考链接:

1. https://github.com/haoel/leetcode/blob/master/algorithms/wordLadder/wordLadder.II.cpp