Use a hash table to solve the problem of anagram grouping.
The key of the map is a sorted string, and the value is the answer
class Solution {
public:
vector<vector<string>> groupAnagrams(vector<string>& strs) {
unordered_map<string,vector<string>> mp;
for(string& str : strs){
string key = str;
sort(key.begin(),key.end());
mp[key].push_back(str);
}
vector<vector<string>> ans;
for(auto it = mp.begin(); it != mp.end(); ++it){
ans.push_back(it->second);
}
return ans;
}
};