leetcode884：兩句話中不常見的詞

使用map存儲每個詞的次數，遍歷兩句話，次數爲1的即爲不常見的詞。

其中判斷是否等於空格的十進制數32，若等於，則說明一個詞結束了。

  vector<string> uncommonFromSentences(string A, string B) {
        map<string,int> mymap;
        int pos = 0 ;
       for(int i = 0; i < A.size(); i ++)
        {
            if(A[i] ==  32 )
            {
                mymap[A.substr(pos, i-pos)]++;
                ///cout <<  A.substr(pos,i)<< endl;
                if((i+1) != A.size())
                     pos = i+1;
            }
            if(i+1 == A.size())
                mymap[A.substr(pos, i-pos+1)]++;
        }
        pos = 0;
        for(int j = 0; j < B.size(); j++)
        {
            if(B[j] == 32)
            {
                mymap[B.substr(pos, j-pos)] ++ ;
                if(j+1 != B.size() )
                    pos = j + 1;
            }
            if(j+1 == B.size())
                mymap[B.substr(pos, j-pos+1)]++;
        }
        vector<string> res;
        map<string,int>::iterator it = mymap.begin();
        for(it = mymap.begin(); it != mymap.end() ; it ++)
        {
          //cout <<  it-> first<<endl;
            if(it->second == 1)
            {
                res.push_back(it->first);
            }
        }
        return res;
    }