Given an array of strings, return all groups of strings that are anagrams.
Note: All inputs will be in lower-case.
NOTE: the return is the list of all anagrams groups, the order does not matter, just put the words in each anagram group into the list.
Analysis:
Count the number of appearance of each char in a word, and generate a uid string like "num[0]/num[1]/....num[25]/", then use hash table to store the words that have the same uid.
Solution:
1 public class Solution { 2 public List<String> anagrams(String[] strs) { 3 List<String> res = new LinkedList<String>(); 4 if (strs.length==0 || strs.length==1) return res; 5 6 Map<String,List<String>> resMap = new HashMap<String,List<String>>(); 7 List<String> uidList = new ArrayList<String>(); 8 9 for (int i=0;i<strs.length;i++){ 10 String uid = getUID(strs[i]); 11 if (resMap.containsKey(uid)) 12 resMap.get(uid).add(strs[i]); 13 else { 14 List<String> list = new LinkedList<String>(); 15 list.add(strs[i]); 16 resMap.put(uid,list); 17 uidList.add(uid); 18 } 19 } 20 21 for (int i=0;i<uidList.size();i++){ 22 List<String> list = resMap.get(uidList.get(i)); 23 if (list.size()>1) res.addAll(list); 24 } 25 26 return res; 27 } 28 29 30 public String getUID(String s){ 31 int[] num = new int[26]; 32 Arrays.fill(num,0); 33 for (int i=0;i<s.length();i++){ 34 char temp = s.charAt(i); 35 int val = temp-'a'; 36 num[val]++; 37 } 38 39 String uid = ""; 40 for (int i=0;i<26;i++) 41 uid+=Integer.toString(num[i])+"/"; 42 43 return uid; 44 45 } 46 47 }