[LeetCode] Word Break II

时间：2014-06-17 21:15:35 阅读：211 评论：0 收藏：0 [点我收藏+]

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.

Return all such possible sentences.

For example, given s = "catsanddog", dict = ["cat", "cats", "and", "sand", "dog"].

A solution is ["cats and dog", "cat sand dog"].

以下时间复杂度过高，没有通过的：

void word_Merge(vector<string> &result,  string s0,int newBegin,  int len,   map<pair<int,int>,string> m)
                 //result存放结果，s0是result存放的值，newBegin是下一个subString开始的下标                                                                       
{
    map<pair<int,int>,string>::iterator iter1;
    if(newBegin == len)
        result.push_back(s0);
    else
    {
       for(iter1 = m.begin(); iter1 != m.end(); iter1++)
       {
           string newS = s0;
           if((*iter1).first.first == newBegin)
           {
               newS += (*iter1).second;
               if((*iter1).first.second != len-1)
                   newS += " ";
                word_Merge(result,newS,(*iter1).first.second+1,len,m);
           }//end if
       }//end for
    }//end if else
}//end word_Merge

class Solution {
public:vector<string> wordBreak(string s, unordered_set<string> &dict) {
      
    map<pair<int,int>,string> m;  //m中存放的是每个子字符串以及对应的开始结束位置<int,int>;
      
      for(string::size_type i=0;i<s.size();i++){
          for(string::size_type j=i;j<s.size();j++){
            string str = s.substr(i,j-i+1);
            if(dict.find(str) != dict.end()){
                m.insert( make_pair( make_pair(i,j),str) );
            }//end if
          }//end for
      }//end for

      vector<string> result;
      int len = s.size();
      for(map<pair<int,int>,string>::iterator iter=m.begin();iter!=m.end();iter++){
         
             if((*iter).first.first == 0){
              string S0 = (*iter).second;
              if((*iter).first.second != len-1)
                  S0 += " ";
               word_Merge(result,S0,(*iter).first.second+1,len,m);
          }//end if
          
      }//end for
   return result;
}//end wordBreak
};

没有用动态规划（dynamic programming），进行很多的递归调用。虽然函数功能没有问题，但通不过很正常。

下面看使用动态规划的方法：

void collect(vector<vector<int>>& mark, int ind, const string& s, string path, vector<string>& result){
             //ind是s中子字符串开始的下标，mark里第ind项存放子字符串结束的下标+1即stop。
        for(int i=0;i<mark[ind].size();i++){
            int stop = mark[ind][i];
            string sub=s.substr(ind,stop-ind);
            string newpath=path+(ind==0?sub:" "+sub);   ////此句写的很简洁
            if(stop==s.length()) 
               result.push_back(newpath);
            else 
               collect(mark,stop,s,newpath,result);
    }//end for
}//end collect
vector<string> wordBreak(string s, unordered_set<string> &dict) {

    vector<vector<int>> mark(s.length(),vector<int>());
    for(int stop=s.length();stop>=0;stop--){    //用两个for循环给mark赋值，中间用了if...continue..大大节省了时间

        if(stop<s.length() && mark[stop].empty())continue;//本句最节省时间！！！
//如果没有以下标stop开始的字符串，就不必有下面for循环中以stop-1结束的字符串了，所以continue
        for(int start=stop-1;start>=0;start--)//以下标start开始，stop结束的字符串，即s中[start,stop)之间;
            if(dict.count(s.substr(start,stop-start)))
                mark[start].push_back(stop);
    }//end for
    vector<string> result;
    collect(mark,0,s,"",result);
    return result;
}//end wordBreak