LeetCode127：Word Ladder II

题目：

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

  [
    ["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：

这道题可真是耗费我老多时间，提交结果要不就是超时，要不就是内存限制，哎，编程能力急待提高啊。

本题是Word Ladder的升级版，可对我这菜鸟才说升的可不止一级。该题主要的解题思路就是BFS+DFS。首先如Word Ladder，构造抽象图，然后利用BFS找到从start到end的最短路径，但寻找最短路径的过程中，我们需要保持最短路径最个节点的父节点信息，以便之后进行所有最短路径结果的重构，重构路径采用的DFS回溯，从end节点开始一直到start节点。好了，不多说了，直接上代码。

实现代码：（注：该代码未能AC，出现内存限制，忘牛人指正，实在不知错在哪里）

#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <unordered_set>
#include <unordered_map>
#include <algorithm>
using namespace std;

/**
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
  [
    ["hit","hot","dot","dog","cog"],
    ["hit","hot","lot","log","cog"]
  ]
Note:
All words have the same length.
All words contain only lowercase alphabetic characters.
*/

class Solution {
public:
    vector<vector<string>> findLadders(string start, string end, unordered_set<string> &dict) {
        vector<vector<string>> resvec;
        if(start.empty() || end.empty() || dict.empty())
            return resvec;
        unordered_map<string, vector<string>> premap;
        //这里需要用到两个vector来模拟两个队列而不是直接用两个队列，是因为需要对队列进行遍历，queue做不到 
        vector<string> squ[2]; 
        squ[0].push_back(start);
        bool qid = false;
        bool finish = false;
        while(!squ[qid].empty())
        {
            squ[!qid].clear();
            vector<string>::iterator iter;
            for(iter = squ[qid].begin(); iter != squ[qid].end(); ++iter)
                dict.erase(*iter);//从dict中删除同一层的所有节点，以免造成循环操作 
            for(iter = squ[qid].begin(); iter != squ[qid].end(); ++iter)//处理同一层节点 
            {
                string curstr = *iter;                
                for(int i = 0; i < curstr.size(); i++)
                {
                    char t = curstr[i];
                    for(char j = 'a'; j <= 'z'; j++)
                    {
                        if(j == curstr[i])
                            continue;
                        curstr[i] = j;
                        if(curstr == end)
                        {
                            finish = true;
                            premap[curstr].push_back(*iter);
                            
                        }                            
                        else if(dict.count(curstr) > 0)
                        {
                            squ[!qid].push_back(curstr);
                            premap[curstr].push_back(*iter);
                        }
                    }
                    curstr[i] = t;
                    
                }                                            
            }
            if(finish)//说明已经处理到了end节点，可以直接break循环，进行结果重构了 
                    break;            
            qid = !qid;//表示将要处理的下一层 
        }
        if(premap.count(end) == 0)//表明end节点的父节点不存在，所有没有到end的转换，直接返回空resvec 
            return resvec;
        vector<string> tmp;
        getResult(resvec, tmp, premap, start, end);
        return resvec;        
    }
    

    
    //DFS 
    void getResult(vector<vector<string> > &resvec, vector<string> &tmp,   
        unordered_map<string, vector<string> > &premap, string &start, string &cur)  
    {  
        tmp.push_back(cur);
        if (cur == start)  
        {  
            resvec.push_back(tmp);  
            reverse(resvec.back().begin(), resvec.back().end());    
        }
        else
        {
            vector<string> v = premap[cur];  
            for (int i = 0; i < v.size(); i++)  
            {   
                getResult(resvec, tmp, premap, start, v[i]);  
            }            
        }  
        tmp.pop_back();  
    }
};


int main(void)
{
    string start("hit");
    string end("cog");
    string strarr[] = {"hot","dot","dog","lot","log"};
    int n = sizeof(strarr) / sizeof(strarr[0]);
    unordered_set<string> dict(strarr, strarr+n);
    Solution solution;
    vector<vector<string>> resvec = solution.findLadders(start, end, dict);
    vector<vector<string>>::iterator iter;
    for(iter = resvec.begin(); iter != resvec.end(); ++iter)
    {
        vector<string> tmp = *iter;
        vector<string>::iterator it;
        for(it = tmp.begin(); it != tmp.end(); ++it)
            cout<<*it<<" ";
        cout<<endl;
    }
    return 0;
}

以下附网上大神AC代码一份：

class Solution {
public:
    vector<vector<string> > findLadders(string start, string end, unordered_set<string> &dict)
    {
        result_.clear();
        unordered_map<string, vector<string>> prevMap;
        for(auto iter = dict.begin(); iter != dict.end(); ++iter)
            prevMap[*iter] = vector<string>();
        vector<unordered_set<string>> candidates(2);
        int current = 0;
        int previous = 1;
        candidates[current].insert(start);
        while(true)
        {
            current = !current;
            previous = !previous;
            for (auto iter = candidates[previous].begin(); iter != candidates[previous].end(); ++iter)
                dict.erase(*iter);
            candidates[current].clear();
            
            for(auto iter = candidates[previous].begin(); iter != candidates[previous].end(); ++iter)
            {
                for(size_t pos = 0; pos < iter->size(); ++pos)
                {
                    string word = *iter;
                    for(int i = 'a'; i <= 'z'; ++i)
                    {
                        if(word[pos] == i)continue;
                        word[pos] = i;
                        if(dict.count(word) > 0)
                        {
                            prevMap[word].push_back(*iter);
                            candidates[current].insert(word);
                        }
                    }
                }
            }
            if (candidates[current].size() == 0)
                return result_;
            if (candidates[current].count(end)) break;
        }
        vector<string> path;
        GeneratePath(prevMap, path, end);
        return result_;
    }
    
private:
    void GeneratePath(unordered_map<string, vector<string>> &prevMap, vector<string>& path, const string& word)
    {
        if (prevMap[word].size() == 0)
        {
            path.push_back(word);
            vector<string> curPath = path;
            reverse(curPath.begin(), curPath.end());
            result_.push_back(curPath);
            path.pop_back();
            return;
        }
        path.push_back(word);
        for (auto iter = prevMap[word].begin(); iter != prevMap[word].end(); ++iter)
            GeneratePath(prevMap, path, *iter);
        path.pop_back();
    }
    vector<vector<string>> result_;
};

代码来源：http://blog.csdn.net/doc_sgl/article/details/13341405