Word Ladder II

leetcode: 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.

解题思路：

1、使用图的广度优先搜索策略（BFS）；

2、由于要返回所有的最短转换序列，使用unordered_map<string , vector<string>>来存储每个节点的前驱节点；

3、使用递归的形式，将unordered_map<string , vector<string>>中存放的节点顺序信息恢复成转换序列。

代码：

class Solution {
private:
        vector> pathes;
        void buildPathes(unordered_map> &traces , vector &path , const string end)
        {
            if(traces[end].size() == 0)
            {
                path.push_back(end);
                vector curPath = path;
                reverse(curPath.begin() , curPath.end());
                pathes.push_back(curPath);
                path.pop_back();
                return;
            }
            vector prevs = traces[end];
            path.push_back(end);
            for(vector::iterator it = prevs.begin() ; it != prevs.end() ; it++)
            {
                buildPathes(traces , path , *it);
            }
            path.pop_back();
        }
public:
    vector> findLadders(string start, string end, unordered_set &dict) {
        pathes.clear();
        dict.insert(start);
        dict.insert(end);
        vector prev;
        unordered_map> traces;
        for(unordered_set::iterator it = dict.begin() ; it != dict.end() ; it++)
        {
            traces[*it] = prev;
        }
        vector> layer(2);
        bool cur = true;
        bool pre = false;
        layer[cur].insert(start);
        while(true)
        {
            cur = !cur;
            pre = !pre;
            for(unordered_set::iterator it = layer[pre].begin(); it != layer[pre].end() ; it++)
            {
                dict.erase(*it);
            }
            layer[cur].clear();
            
            for(unordered_set::iterator it = layer[pre].begin(); it != layer[pre].end() ; it++)
            {
                for(int i = 0 ; i < (*it).size() ; i++)
                {
                    string word = *it;
                    int stop = word[i] - 'a';
                    for(int c = (stop + 1)%26 ; c != stop ; c = (c + 1)%26)
                    {
                        word[i] = c + 'a';
                        if(dict.find(word) != dict.end())
                        {
                            traces[word].push_back(*it);
                            layer[cur].insert(word);
                        }
                    }
                }
            }
            if(layer[cur].size() == 0)
                return pathes;
            if(layer[cur].count(end))
                break;
        }
        vector path;
        buildPathes(traces, path, end);
        return pathes;
    }
};

疑点：根据unordered_map<string , vector<string>>中的节点信息，从end节点开始，递归往前推导，得出来的转换序列是全部转换序列，还是所有最短转换序列？答案肯定是返回所有的最短转换序列，首先可以这样想，在广度优先搜索的时候，Layer[pre]和Layer[cur]像剥洋葱一样，一层一层的向前推进，直到搜索到单词end的时候结束，那么从end再一层一层返回，得到的也会是最短的转换序列。换一种思路，假设在原单词表中，单词a可以直接到单词x，即a -> x，也可以经过单词b到单词x，即a -> b -> c ->x。当{a}为当前所搜索的层次时，记录x的前驱是a，b的前驱是a，所以，接下来搜索的一层是{b ， x}，然后在单词表中删除b、x，由此可以得出，b -> c -> x这个路径是记录不下来的。所以，广度优先搜索记录下来的路径是最短路径。