[leetcode-49-Group Anagrams]

Given an array of strings, group anagrams together.
For example, given: ["eat", "tea", "tan", "ate", "nat", "bat"],
Return:
[
["ate", "eat","tea"],
["nat","tan"],
["bat"]
]

思路：

用map和set，set用来保存属于同一个anagram的string，map，用来对应不同的anagram。

vector<vector<string> > groupAnagrams(vector<string>& strs)
{
  vector<vector<string> >group;
  if(strs.size()==0)return group;
  map<string,set<string> >m;
  for(int i=0;i<strs.size();i++)
  {
    string temp = strs[i];
    sort(temp.begin(),temp.end());
    m[temp].insert(strs[i]);
  }
  map<string,set<string> >::iterator it;
  for(it = m.begin();it!=m.end();it++)
  {
    vector<string> anagram;
    set<string>::iterator setit;
    for(setit = it->second.begin();setit!=it->second.end();setit++)
    {
      anagram.push_back(*setit);
    }
    group.push_back(anagram);
  }
  return group;  
}

如上，总是不能通过测试用例["",""],而且写的感觉也啰嗦，如下是参考的别人的：

vector<vector<string>> groupAnagrams(vector<string>& strs) {
 unordered_map<string, multiset<string>> mp;
        for (string s : strs) {
            string t = s; 
            sort(t.begin(), t.end());
            mp[t].insert(s);
        }
        vector<vector<string>> anagrams;
        for (auto m : mp) { 
            vector<string> anagram(m.second.begin(), m.second.end());
            anagrams.push_back(anagram);
        }
        return anagrams;
   
}

参考：

https://discuss.leetcode.com/topic/21038/10-lines-76ms-easy-c-solution-updated-function-signature