对每个string做循环，记录里面每种字母的数量，然后加入到map中，最后map中相同键值的就是异位词。

class Solution {
public:
vector<vector<string>> groupAnagrams(vector<string> &strs)
{
int n = strs.size();
if (!n)
return {};
unordered_map<string, vector<string>> m;
for (int i = 0; i < n; i++)
{
string count = "00000000000000000000000000";
for (int k = 0; k < strs[i].size(); k++)
count[strs[i][k] - 'a'] += 1;
//else
m[count].push_back(strs[i]);
}
vector<vector<string>> ret;
vector<string> temp;
for(auto it=m.begin();it!=m.end();it++)
ret.push_back(it->second);

return ret;
}
};