题目

Given an array of strings, group anagrams together.

For example, given: ["eat", "tea", "tan", "ate", "nat", "bat"],
Return:

1
2
3
4
5
[
  ["ate", "eat","tea"],
  ["nat","tan"],
  ["bat"]
]

大意

找出字符相同的所有字符串,归为一组.

答案

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        int len = strs.size();
        vector<vector<string> > res;
        map<string,vector<string>> maps;
        for(int i=0;i<len;i++)
        {
            string s =strs[i];
            sort(s.begin(),s.end());
            maps[s].push_back(strs[i]);
        }
        for (map<string, vector<string> >::iterator iter = maps.begin(); iter != maps.end(); iter++)
            res.push_back(iter->second);
        return res;
    }
};

思路

穷举会超时,所以这道题要使用哈希表,每个字符串的每个单词进行排序,排序完成后放到相应的map中.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
//Java 版本
public class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        if (strs == null || strs.length == 0) return new ArrayList<List<String>>();
        Map<String, List<String>> map = new HashMap<String, List<String>>();
        for (String s : strs) {
            char[] ca = s.toCharArray();
            Arrays.sort(ca);
            String keyStr = String.valueOf(ca);
            if (!map.containsKey(keyStr)) map.put(keyStr, new ArrayList<String>());
            map.get(keyStr).add(s);
        }
        return new ArrayList<List<String>>(map.values());
    }
}