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;
}