Leetcode 49 Group Anagrams

没做出来，自己的算法超时

class Solution {public:    typedef map<char, int>::iterator MIT;    vector<vector<string>> groupAnagrams(vector<string>& strs) {        // mark the work        vector<vector<string> > ret;        vector<map<char,int> > maps;        size_t n = strs.size();        for (size_t i = 0; i < n; ++i) {            string str = strs[i];            size_t strlen = str.size();            map<char, int> tmp;            for (size_t j = 0; j < strlen; ++j) {                ++tmp[str[j]];            }                    size_t k = 0;            for (; k != maps.size(); ++k) {                             if (maps[k].size() != tmp.size())  continue;                MIT itmp = tmp.begin();                for (; itmp != tmp.end(); ++itmp) {                    MIT imaps = maps[k].find(itmp->first);                    if (maps[k].end() == imaps || imaps->second != itmp->second)                        break;                                        }                if (itmp != tmp.end())                     continue;                else {                    ret[k].push_back(str);                    break;                }            }            // 没找到            if (k == maps.size()) {                ret.push_back(vector<string>(1, str));                maps.push_back(tmp);            }        }          for (size_t m = 0; m <ret.size(); ++m) {            sort(ret[m].begin(), ret[m].end());        }        return ret;    }};

参考后

class Solution {public:    vector<vector<string>> groupAnagrams(vector<string>& strs) {        unordered_map <string, multiset<string> > mp;        for (auto str : strs) {       // for (size_t i = 0; i != strs.size(); ++i) {            //string str = strs[i];           string keystr = sortStr(str);            // sort(keystr.begin(), keystr.end());            mp[keystr].insert(str);        }        vector<vector<string> > ret;        for (auto m : mp) {            ret.push_back(vector<string>(m.second.begin(), m.second.end()));        }        return ret;    }private:// sort 时间复杂度 O(n*logn)// sortStr 时间复杂度 o(n)    string sortStr(string s) {        string ret;        vector<int> flag(26, 0);        for (auto ch : s) {            ++flag[ch - 'a'];        }        for (size_t i = 0; i < 26; ++i) {            ret += string(flag[i], 'a' + i);        }                return ret;    }};