Anagrams
Given an array of strings, return all groups of strings that are anagrams.

Note: All inputs will be in lower-case.

思路:

建Hashtable,用排序过的string作为key,它的anagram作为ArrayList

这道题之前用暴力写的O(N^2)的TLE了,改用Hashtable来写
题目的意思是给一个String数组,找出其中由相同字母组成的单词。
例如:
S = ["abc", "bca", "bac", "bbb", "bbca", "abcb"]
答案为:
["abc", "bca", "bac", "bbca", "abcb"]
只有"bbb"没有相同字母组成的单词。

ref: http://blog.csdn.net/fightforyourdream/article/details/14217985

 1 public class Solution {
 2     public List<String> anagrams(String[] strs) {
 3         List<String> ret = new ArrayList<String>();
 4         
 5         if (strs == null) {
 6             return ret;
 7         }
 8         
 9         HashMap<String, List<String>> map = new HashMap<String, List<String>>();
10         
11         int len = strs.length;
12         for (int i = 0; i < len; i++) {
13             String s = strs[i];
14             
15             // Sort the string.            
16             char[] chars = s.toCharArray();
17             Arrays.sort(chars);
18             String strSort = new String(chars);   
19             
20             // Create a ArrayList for the sorted string.            
21             if (!map.containsKey(strSort)) {
22                 map.put(strSort, new ArrayList<String>());
23             }
24             
25             // Add a new string to the list of the hashmap.
26             map.get(strSort).add(s);
27         }
28         
29         // go through the map and add all the strings into the result.
30         for (Map.Entry<String, List<String>> entry: map.entrySet()) {
31             List<String> list = entry.getValue();
32             
33             // skip the entries which only have one string.
34             if (list.size() == 1) {
35                 continue;
36             }
37             
38             // add the strings into the list.
39             ret.addAll(list);
40         }
41         
42         return ret;
43     }
44 }
View Code

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