Substring with Concatenation of All Words

题目：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 words exactly
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：

定义两个hash表，一个就是传统的把所有的输入进去，对应代号。

本题的一个技巧就是每次从一个位置开始判断，然后再最外面写一个从当前位置判断再这个 start  能否成功的函数：

函数的技巧就是从start位置截取每一个单词的长度，比对是否出现过，出现的次数不能超过1次。

代码：

class Solution {
public:
vector<int> findSubstring(string s, vector<string>& words) {
vector<int> res;
if(words.empty())   return res;
int wordSize=words[0].size();
int totalWords=words.size();
int totalLen=wordSize*totalWords;

if(s.size()<totalLen)   return res;

unordered_map<string,int>wordCount;
for(int i=0;i<totalWords;i++){
wordCount[words[i]]++;
}

for(int i=0;i<=s.size()-totalLen;i++){
if(checkSubstring( s,i, wordCount, wordSize, totalWords))   res.push_back(i);
}
return res;
}

bool checkSubstring(string &s,int start,unordered_map<string,int>&wordCount,int wordSize,int totalWords){
if(s.size()-start+1<wordSize*totalWords)    return false;
unordered_map<string,int>wordFound;

for(int i=0;i<totalWords;i++){
string tmp=s.substr(start+i*wordSize,wordSize);
if(wordCount[tmp]==0)   return false;
wordFound[tmp]++;
if(wordFound[tmp]>wordCount[tmp])   return false;
}

return true;
}
};