49. Group Anagrams

class Solution 
{
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) 
    {
        vector<vector<string>> res;
        if(strs.empty())
            return res;
        unordered_map<string,vector<string>> svmap;
        for(string &s : strs)
            svmap[sortstring(s)].push_back(s);
        for(auto p : svmap)
            res.push_back(p.second);
        return res;
    }
    
    string sortstring(string & st)
    {
        string t=st;
        sort(t.begin(),t.end());
        return t;
    }
};

这里利用了乱序序列排序后是相等的这一点，把所有字符串排序，然后用map来添加至容器，很机智