Given an array of strings, group anagrams together.

For example, given: ["eat", "tea", "tan", "ate", "nat", "bat"], Return:

[  ["ate", "eat","tea"],  ["nat","tan"],  ["bat"]]my initial answer:
class Solution {public:    vector<vector<string>> groupAnagrams(vector<string>& strs) {        set<char> charSet;        vector<int> indexSet;        map<set<char>,vector<int>> myMap;        vector<vector<string>> result;        vector<string> temPResult;        for(int i = 0; i < strs.size(); i ++){            string tempStr = strs[i];            set<char> tempCharSet;            vector<int> tempIndexSet;            for(int j = 0; j < tempStr.size(); j ++){                tempCharSet.insert(tempStr[i]);            }            map<set<char>,vector<int>>::iterator it = myMap.find(tempCharSet);            if(it == myMap.end()){                tempIndexSet.push_back(i);                myMap.insert(make_pair(tempCharSet,tempIndexSet));            }            else ((*it).second).push_back(i);        }        map<set<char>,vector<int>>::iterator it = myMap.begin();        while(it != myMap.end()){            indexSet = (*it).second;            for(int i = 0; i < indexSet.size(); i ++){                tempResult.push_back(strs[i]);            }            result.push_back(tempResult);            tempResult.clear();            it ++;        }        return result;    }}; 但是在最長(zhǎng)的那個(gè)case，超時(shí)了，后來(lái)看了別人的算法，改進(jìn)了。
class Solution {public:    vector<vector<string>> groupAnagrams(vector<string>& strs) {        map<string,vector<string>> myMap;        for(int i = 0; i < strs.size(); i ++){            string temp = strs[i];            sort(temp.begin(),temp.end());            myMap[temp].push_back(strs[i]);        }        vector<vector<string>> result;                map<string,vector<string>>::iterator it = myMap.begin();        vector<string> temp;        while(it != myMap.end()){            temp = (*it).second;            result.push_back(temp);            it ++;        }        return result;    }};