예제 #1
0
inline bool operator==(const std::tr1::unordered_set<size_t> &a, const std::tr1::unordered_set<size_t> &b) {
  if (a.size() != b.size()) return false;

  std::tr1::unordered_set<size_t>::const_iterator cit = a.begin();
  while (a.end() != cit) {
    if (b.end() == b.find(*cit))
      return false;
    cit++;
  }
 
  return true;
}
예제 #2
0
파일: main.cpp 프로젝트: 540485909/leetcode
vector<bool> wordBreak1(string s, std::tr1::unordered_set<string> &dict) {
       vector<bool> wordB(s.length()+1,false);       //wordB stands for the string from 0 to i can be segument

        wordB[0]=true;
        for(int i=1;i<s.length()+1;i++)
        {
            for(int j=i-1;j>=0;j--)
            {
                if(wordB[j]&&dict.find(s.substr(j,i-j))!=dict.end())
                    {
                        wordB[i]=true;
                        break;
                    }
            }
        }
        return wordB;


}
예제 #3
0
파일: main.cpp 프로젝트: 540485909/leetcode
void dfs(int start,int n,string s,string cur,std::tr1::unordered_set<string> &dict,vector<bool> wordB)
{
    if(start==n)
    {
        result.push_back(cur);
        return ;
    }
    for(int i=start;i<n;i++)
    {
        string t=s.substr(start,i+1-start);
        if(wordB[i+1]&&dict.find(t)!=dict.end())
        {
            int c=cur.size();//used to record the current element ,will be used to keep the size like the pop operation,all is caused by " "signal
            cur+=t;
            if(i<n-1)
            cur.push_back(' ');//the last element do not need to add a space symbol
            dfs(i+1,n,s,cur,dict,wordB);
            cur.resize(c);//being used to back to the previous condition
        }
    }
}