LeetCode | Substring with Concatenation of All Words（链接所有单词的子串）_you are given a string, s, and a list of words, wo

You are given a string, S, and a list of words, L, that are all of the same length. Find all starting indices of substring(s) in S that is a concatenation of each word in L exactly once and without any intervening characters.

For example, given:
S: "barfoothefoobarman"
L: ["foo", "bar"]

You should return the indices: [0,9].
(order does not matter).

题目解析：

这道题纠结了好久……由于对c++不熟悉，一直想用c来实现，结果就太麻烦了，不知道哪位大神用c成功实现，忘分享……

用c++的map就相当容易了，先将L中的字符串在map中建立映射。然后S从前往后一次以某一个为起始字符判断长度为l_size*word_size的连续字符串是否都在map中。当然为了防止L中有重复的字符串出现，另设一个counting，来记录子串的个数，当个数超过word_count[word]的时候，也退出，遍历S中的下一个字符。

class Solution {
public:
    vector<int> findSubstring(string S, vector<string> &L) {
        int l_size = L.size();
        vector<int> res;
        if(l_size <= 0)
            return res;
 
        map<string,int> word_count; //记录L字符串中每一个元素出现的个数
        int word_size = L[0].size();
        for(int i = 0;i < l_size;i++)
            ++word_count[L[i]];
        map<string,int> counting;   //记录查找过程中的个数
        for(int i = 0;i <= (int)S.size()-l_size*word_size;i++){
            counting.clear();
            int j;
            for(j = 0;j < l_size;j++){
                string word = S.substr(i+j*word_size,word_size);
                if(word_count.find(word) != word_count.end()){
                    ++counting[word];
                    if(counting[word] > word_count[word])
                        break;
                }else
                    break;
            }
            if(j == l_size)
                res.push_back(i);
        }
        return res;
    }
};