力扣 | 49. 字母异位词分组

这里使用HashMap

力扣 | 49. 字母异位词分组_第1张图片

Java

package _49;

import java.util.*;

public class Problem_49_GroupAnagrams {
    public List<List<String>> groupAngrams(String [] strs){
        Map<String,List<String>> map = new HashMap<>();
//        int [] arr = new int[]{};
        for(String s : strs){
            char[] chars = s.toCharArray();
            Arrays.sort(chars);//排序之后的字母异位词都是一样,因此可以作为map的key
            String key = String.valueOf(chars);
            if(!map.containsKey(key)) map.put(key,new ArrayList<>());
            map.get(key).add(s);
        }

        return new ArrayList<>(map.values());//

    }
  }


public List<List<String>> groupAnagrams1(String[] strs) {
        Map<String, List<String>> map = new HashMap<>();
        for (String s : strs) {
            char[] chars = s.toCharArray();
            int[] count = new int[26];
            for (char c : chars) count[c - 'a']++;
            String key = Arrays.toString(count);//int: 32位有符号整数,范围是-2^31到2^31-1。
            if (!map.containsKey(key)) map.put(key, new ArrayList<>());
            map.get(key).add(s);
        }
        return new ArrayList<>(map.values());
    }

C++

class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        unordered_map<string, vector<string>> map;
        for (string& str : strs) {
            vector<int> count(26);
            for (char c : str) count[c - 'a']++;
            string key = "";
            for (int i = 0; i < 26; i++) key += count[i];
            map[key].emplace_back(str);
        }
        vector<vector<string>> ans;
        for (auto it = map.begin(); it != map.end(); ++it) {
            ans.emplace_back(it->second);
        }
        return ans;
    }

    vector<vector<string>> groupAnagrams1(vector<string>& strs) {
        unordered_map<string, vector<string>> map;
        for (string& str : strs) {
            string key = str;
            sort(key.begin(), key.end());
            map[key].emplace_back(str);
        }
        vector<vector<string>> ans;
        for (auto it = map.begin(); it != map.end(); ++it) {
            ans.emplace_back(it->second);
        }
        return ans;
    }
};

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