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());
}
}