49. 字母异位词分组(group-anagrams)

字母异位词分组(group-anagrams)

给定一个字符串数组，将字母异位词组合在一起。字母异位词指字母相同，但排列不同的字符串。
示例:

输入: ["eat", "tea", "tan", "ate", "nat", "bat"],
输出:
[
  ["ate","eat","tea"],
  ["nat","tan"],
  ["bat"]
]

说明：
所有输入均为小写字母。
不考虑答案输出的顺序。

代码与思路

方法一 排序数组分类

cpp代码

#include <stdio.h>

#include <vector>
#include <string>
#include <map>
#include <algorithm>

class Solution {
public:
    std::vector<std::vector<std::string> > groupAnagrams(
			std::vector<std::string>& strs) {
		std::map<std::string, std::vector<std::string> > anagram;
		//内部进行排序的各个单词为key，以字符串向量(vector<string>)为
		//value,存储各个字符数量相同的字符串(anagram)
		std::vector<std::vector<std::string> > result;//存储最终的结果		
		for (int i = 0; i < strs.size(); i++){//遍历各个单词
			std::string str = strs[i];
			std::sort(str.begin(), str.end());//对str内部进行排序
			if (anagram.find(str) == anagram.end()){//若无法在哈希表中找到str
				std::vector<std::string> item;//设置一个空的字符串向量
				anagram[str] = item;//以排序后的strs[i]作为key
			}
			anagram[str].push_back(strs[i]);//在对应的字符串向量中push结果
		}
		//遍历哈希表，将哈希表的value push进入最终结果
		std::map<std::string, std::vector<std::string> > ::iterator it;
		for (it = anagram.begin(); it != anagram.end(); it++){
			result.push_back((*it).second);
		}
    	return result;
    }
};

int main(){
	std::vector<std::string> strs;
	strs.push_back("eat");
	strs.push_back("tea");
	strs.push_back("tan");
	strs.push_back("ate");
	strs.push_back("nat");
	strs.push_back("bat");
	Solution solve;
	std::vector<std::vector<std::string> > result 
		= solve.groupAnagrams(strs);
	for (int i = 0; i < result.size(); i++){
		for (int j = 0; j < result[i].size(); j++){
			printf("[%s]", result[i][j].c_str());
		}
		printf("\n");
	}	
	return 0;
}

java代码

class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        if (strs.length == 0) return new ArrayList();
        Map<String, List> ans = new HashMap<String, List>();
        for (String s : strs) {
            char[] ca = s.toCharArray();
            Arrays.sort(ca);
            String key = String.valueOf(ca);
            if (!ans.containsKey(key)) ans.put(key, new ArrayList());
            ans.get(key).add(s);
        }
        return new ArrayList(ans.values());
    }
}

方法二 按计数分类

cpp代码

#include <stdio.h>

#include <vector>
#include <string>
#include <map>

void change_to_vector(std::string &str, std::vector<int> &vec){
	for (int i = 0; i < 26; i++){
		vec.push_back(0);
	}
	for (int i = 0; i < str.length(); i++){
		int cc = str[i] - 'a';
		vec[cc]++;
	}
}

class Solution {
public:
    std::vector<std::vector<std::string> > groupAnagrams(
			std::vector<std::string>& strs) {
		std::map<std::vector<int>, std::vector<std::string> > anagram;
		std::vector<std::vector<std::string> > result;
		for (int i = 0; i < strs.size(); i++){
			std::vector<int> vec;
			change_to_vector(strs[i], vec);
			if (anagram.find(vec) == anagram.end()){
				std::vector<std::string> item;
				anagram[vec] = item;
			}
			anagram[vec].push_back(strs[i]);
		}
		std::map<std::vector<int>,
			std::vector<std::string> > ::iterator it;
		for (it = anagram.begin(); it != anagram.end(); it++){
			result.push_back((*it).second);
		}
    	return result;
    }
};

int main(){
	std::vector<std::string> strs;
	strs.push_back("eat");
	strs.push_back("tea");
	strs.push_back("tan");
	strs.push_back("ate");
	strs.push_back("nat");
	strs.push_back("bat");
	Solution solve;
	std::vector<std::vector<std::string> > result = solve.groupAnagrams(strs);
	for (int i = 0; i < result.size(); i++){
		for (int j = 0; j < result[i].size(); j++){
			printf("[%s]", result[i][j].c_str());
		}
		printf("\n");
	}
	return 0;
}

java代码

class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        if (strs.length == 0) return new ArrayList();
        Map<String, List> ans = 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('#');
                sb.append(count[i]);
            }
            String key = sb.toString();
            if (!ans.containsKey(key)) ans.put(key, new ArrayList());
            ans.get(key).add(s);
        }
        return new ArrayList(ans.values());
    }
}