30. Substring with Concatenation of All Words (String; HashTable)

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

For example, given:
s: "barfoothefoobarman"
words: ["foo", "bar"]

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

思路: 判断一个值是否包含在一个数组中，首先应该想到将这个数组中的元素放入HashTable，否则每次查找都需要O(n)的时间复杂度。

时间复杂度：O(n*size)，其中n为s的长度，size是指数组words包含多少个元素。当words元素不多的时候，我们可以说时间复杂度是线性的O(n)

class Solution {
public:
    vector<int> findSubstring(string s, vector<string>& words) {
        size = words.size();
        sLen = s.length();
        wLen = words[].length();
        wordsLen = wLen * size;
        for(i = ; i < size; i++){
            word_counter[words[i]]++;
        }

        i = ;
        while(i+wordsLen<=sLen){
            for(j = ; j < size; j++){
                cmpStr = s.substr(i+j*wLen, wLen);
                if(word_counter.find(cmpStr)==word_counter.end()){ //不在words中，不符合
                    break;
                }

                counting[cmpStr]++;
                if(counting[cmpStr]>word_counter[cmpStr]){ //出现的次数多过words中的次数，不符合
                    break;
                }
            }
            if(j==size){//找到了一个符合的结果
                ret.push_back(i);
            }
            counting.clear();
            i++;
        }
        return ret;
    }
private:
    string cmpStr;
    vector<int> ret;
    map<string,int> word_counter;
    map<string,int> counting;
    int size; //number of words
    int sLen;
    int wLen;
    int wordsLen;
    int i;
    int j;
};