30. Substring with Concatenation of All Words

vector<int> findSubstring(string s, vector<string>& words) {    if (words.empty())        return vector<int>();    vector<int> ret;    //记录所给words中每个单词的出现次数    map<string, int> word_count;    //每个单词的长度相同    int word_size = strlen(words[0].c_str());    int word_nums = words.size();    //所给匹配字符串的长度    int s_len = strlen(s.c_str());    for (int i = 0; i < word_nums; i++)        ++word_count[words[i]];    int i, j;    map<string, int> temp_count;    for (i = 0; i < s_len - word_nums*word_size + 1; ++i)    {        temp_count.clear();        for (j = 0; j < word_nums; j++)        {            //检验当前单词是否属于words以及出现的次数是否一致            string word = s.substr(i + j*word_size, word_size);            if (word_count.find(word) != word_count.end())            {                ++temp_count[word];                //如果出现的次数与words不一致，则返回错误                if (temp_count[word] > word_count[word])                    break;            }//if            else {                break;            }//else                         }//for         //所有words内的单词,在i起始位置都出现，则将下标i存入结果的vector中        if (j == word_nums)        {            ret.push_back(i);        }//if    }//for    return ret;}