30. Substring with Concatenation of All Words (String, Map)

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[0].length();
wordsLen = wLen * size;
for(i = 0; i < size; i++){
word_counter[words[i]]++;
}

i = 0;
while(i+wordsLen<=sLen){
for(j = 0; 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;
};