Algorithms to Group Words by Anagrams

  • 时间：2020-10-11 15:48:46
  • 分类：网络文摘
  • 阅读：81 次

Given an array of strings, group anagrams together.

Example:
Input: [“eat”, “tea”, “tan”, “ate”, “nat”, “bat”],
Output:

1
2
3
4
5

[
  ["ate","eat","tea"],
  ["nat","tan"],
  ["bat"]
]
[
  ["ate","eat","tea"],
  ["nat","tan"],
  ["bat"]
]

Note:
All inputs will be in lowercase.
The order of your output does not matter.

Group Anagrams by using Hash Key

As the words are all lower-case, we can count the frequency of each letter using a static array (e.g. int[26]), thus O(1) constant space. Then we can compute the key for such occurrence.

Then, we group words by same key, at last we push the values one by one to the result array/vector.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26

class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        vector<vector<string>> result;
        unordered_map<string, vector<string>> data;
        for (const auto &n: strs) {
            data[getKey(n)].push_back(n);
        }
        for (auto it = data.begin(); it != data.end(); ++ it) {
            result.push_back(it->second);
        }
        return result;
    }
private:
    string getKey(const string &s) {
        int cnt[26] = {};
        for (const auto &n: s) {
            cnt[n - 97] ++;
        }
        string r = "";
        for (int i = 0; i < 26; ++ i) {
            r = r + std::to_string(cnt[i]) + ",";
        }
        return r;
    }
};
class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        vector<vector<string>> result;
        unordered_map<string, vector<string>> data;
        for (const auto &n: strs) {
            data[getKey(n)].push_back(n);
        }
        for (auto it = data.begin(); it != data.end(); ++ it) {
            result.push_back(it->second);
        }
        return result;
    }
private:
    string getKey(const string &s) {
        int cnt[26] = {};
        for (const auto &n: s) {
            cnt[n - 97] ++;
        }
        string r = "";
        for (int i = 0; i < 26; ++ i) {
            r = r + std::to_string(cnt[i]) + ",";
        }
        return r;
    }
};

This approach takes O(N) space as we need a hash map to store the occurence key and the corresponding group of words. The time requirement is O(NM) where M is the average length of the words and N is the length of the word list.

Group Anagrams by Sorting

Anagrams are the same if we sort them. Thus, we can sort each string, and use a hash map to push the same Anagrams to their groups.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19

class Solution {
public:   
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        vector<vector<string>> r;
        unordered_map<string, int> cached;
        for (string n: strs) {
            string t = n;
            sort(n.begin(), n.end());
            if (cached.find(n) != cached.end()) {
                int k = cached[n];
                r[k].push_back(t);
            } else {
                r.push_back({t});
                cached[n] = r.size() - 1;
            }
        }        
        return r;
    }
};
class Solution {
public:   
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        vector<vector<string>> r;
        unordered_map<string, int> cached;
        for (string n: strs) {
            string t = n;
            sort(n.begin(), n.end());
            if (cached.find(n) != cached.end()) {
                int k = cached[n];
                r[k].push_back(t);
            } else {
                r.push_back({t});
                cached[n] = r.size() - 1;
            }
        }        
        return r;
    }
};

This approach takes O(N) space and O(N.MLog(M)) time where N is the input list size and M is the average length of the words.

–EOF (The Ultimate Computing & Technology Blog) —

推荐阅读：
成长是一种幸福　 六一作文400字　 扬州作文　 忍，为生命着色　 父与子　 儿童节游园活动作文　 水池的四个角上各有一棵树画图数学题　 数学题：一家四口人要一起吃饭　 怎样加一笔变成五个水杯　 5号楼共75户住宅　
评论列表
添加评论