Given an array of strings, group anagrams together.
- Example:
Input: [“eat”, “tea”, “tan”, “ate”, “nat”, “bat”],
Output:
[
[“ate”,“eat”,“tea”],
[“nat”,“tan”],
[“bat”]
]
Note:
- All inputs will be in lowercase.
- The order of your output does not matter.
解法一
把每个字符串按字母大小排序,使用map来保存排序后的字符串与一组异位词的映射,遍历完数组后将map中每个key的异位词组加入res即可
public List<List<String>> groupAnagrams(String[] strs) {
List<String> list=new ArrayList<String>();
HashMap<String,List<String>> map=new HashMap<String,List<String>>();
for(int i=0;i<strs.length;i++)
{
char[] c=strs[i].toCharArray();
Arrays.sort(c);
String t=String.valueOf(c);
if(!map.containsKey(t))
map.put(t, new ArrayList<String>());
map.get(t).add(strs[i]);
}
return new ArrayList(map.values());
}
Runtime: 11 ms, faster than 99.11% of Java online submissions for Group Anagrams.
Memory Usage: 41.8 MB, less than 86.15% of Java online submissions for Group Anagrams.
解法二
将字母与0~25对应起来,用一个数组存储每个字符串内每个字母的数量,然后把数组转化为一个字符串作为map里的key
public List<List<String>> groupAnagrams(String[] strs){
List<List<String>> res=new ArrayList<List<String>>();
if(strs==null)
return res;
Map<String,List> map=new HashMap<String, List>();
int[] count=new int[26];
for(String s:strs)
{
Arrays.fill(count, 0);
for(char c:s.toCharArray())
count[c-'a']++;
StringBuilder sb=new StringBuilder("");
for(int i=0;i<26;i++)
sb.append(count[i]);
if(!map.containsKey(sb.toString()))
map.put(sb.toString(), new ArrayList());
map.get(sb.toString()).add(s);
}
return new ArrayList(map.values());
}
Runtime: 22 ms, faster than 34.15% of Java online submissions for Group Anagrams.
Memory Usage: 43.3 MB, less than 50.87% of Java online submissions for Group Anagrams.