LEETCODE - 1181【前后拼接】

class Solution {
public:
    string gethead(string str){//获取头单词
        string ret = "";
        int strlen = str.length();
        for(int i = 0; i < strlen; ++i){
            if(str[i] == ' '){
                break;
            }
            ret += str[i];
        }
        return ret;
    }
    string gettail(string str){//获取尾单词
        int strlen = str.length();
        int index = 0;
        for(int i = strlen - 1; i >= 0; i--){
            if(str[i] == ' '){
                index = i;
                break;
            }
        }
        if(index == 0){
            return str;
        }
        return str.substr(index+1,strlen);;
    }
    string deletesubstr(string str,string substr){//删除重复的单词【为了拼接】；只能删除第一个单词，不能删除最后一个单词【与空格有关】
        string strT = str;
        int pos = strT.find(substr);
        if (pos >-1)
        {
            strT.erase(pos,substr.size());
        }
        return strT;
    }
    vector<string> beforeAndAfterPuzzles(vector<string>& phrases) {
        int phraseslen = phrases.size();
        vector<string> head;
        vector<string> tail;
        for(int i = 0; i < phraseslen; ++i){//获取头尾词的数组
            head.push_back(gethead(phrases[i]));
            tail.push_back(gettail(phrases[i]));
        }
        vector<string> ret;
        for(int j = 0; j < phraseslen; ++j){//尾部
            for(int k = 0; k < phraseslen; ++k){
                if(j != k && tail[j] == head[k]){//如果某个子串的头等于另一个子串的尾
                    ret.push_back(phrases[j] + deletesubstr(phrases[k],head[k]));//把他拼接起来放在返回容器中
                    cout << phrases[j] + deletesubstr(phrases[k],head[k]) << endl;
                }
            }
        }
        sort(ret.begin(), ret.end());
        ret.erase(unique(ret.begin(), ret.end()), ret.end());//去重
        return ret;
    }
};