[Anagram][Hash Table] Group Anagram

1. Example
anagram means after sorted they have the same string
Collections.sort(item);
Group anagrams together

s= ["eat", "tea", "tan", "ate", "nat", "bat"]
Return
[
["ate", "eat", "tea"]
["nat", "tan"],
["bat"]
]


2. Implementation
Q1: Return List<List<String>>?
A1: since the size of string array is dynamic

Input:["tea","and","ate","eat","den"]
Output:[["tea","ate","eat"],["and"],["den"]]
Expected:[["den"],["and"],["ate","eat","tea"]]
NOTE: each inner list's elements must follow the lexicographic order 
==> Collections.sort(item);
NOTE:
for (List<String> value: map.values()) {

// //Time :O(m*n) or O(m+n) , m is length of string and n is number of string in string array
//Time:O(n* mlogm), n is size of string array, m is the length of each string, assume they are the same
public List> groupAnagrams(String[] strs)
{




      List> res = new List>();
      if (strs == null || strs.length == 0)
          return res;




      
      //HashTable> table = new HashTable> ();
      HashMap<String, List<String>> map = new HashMap<String, List<String>>();




      for (int i = 0; i < strs.length; i++)
      {

           


           char[] charArr = strs[i].toCharArray();
           Arrays.sort(charArr);
           String str = new String(charArr);

           



           //if ( table.containsKey() )
           if (map.containsKey(str))
           {
               //table.get(key).add( strs[i]);
               map.get(str).add( strs[i] );
           }
           else 
           {
               //for (int j = 0 ; j< strs[i];j++)
               ///{
                //(int) (strs[i].charAt(j)-'a')
              // }
              List<String> item = new ArrayList<String>();
              item.add(strs[i]);
              //table.put(, item);
              map.put(str, item)
           }




        
      }






      //NOTE:return all the map values,(map.keySet())
      for (List value: map.values())
      {
           List<String> item = new ArrayList<String>(value);





           Collections.sort(item);





           if (item.size() >0)
              res.add(item);
          
      }



      return res;

    

}

3. Similar Ones
(E) Valid Anagram
(E) Group Shifted Strings