давайте возьмем вопрос: учитывая две строки s и t, напишите функцию, чтобы определить, является ли t анаграммой s.
Например,
s = "анаграмма", t = "нагарам", верните истину.
s = "крыса", t = "машина", возврат false.
Метод 1 (с использованием HashMap):
public class Method1 {
public static void main(String[] args) {
String a = "protijayi";
String b = "jayiproti";
System.out.println(isAnagram(a, b ));// output => true
}
private static boolean isAnagram(String a, String b) {
Map<Character ,Integer> map = new HashMap<>();
for( char c : a.toCharArray()) {
map.put(c, map.getOrDefault(c, 0 ) + 1 );
}
for(char c : b.toCharArray()) {
int count = map.getOrDefault(c, 0);
if(count == 0 ) {return false ; }
else {map.put(c, count - 1 ) ; }
}
return true;
}
}
Метод 2:
public class Method2 {
public static void main(String[] args) {
String a = "protijayi";
String b = "jayiproti";
System.out.println(isAnagram(a, b));// output=> true
}
private static boolean isAnagram(String a, String b) {
int[] alphabet = new int[26];
for(int i = 0 ; i < a.length() ;i++) {
alphabet[a.charAt(i) - 'a']++ ;
}
for (int i = 0; i < b.length(); i++) {
alphabet[b.charAt(i) - 'a']-- ;
}
for( int w : alphabet ) {
if(w != 0 ) {return false;}
}
return true;
}
}
Метод 3:
public class Method3 {
public static void main(String[] args) {
String a = "protijayi";
String b = "jayiproti";
System.out.println(isAnagram(a, b ));// output => true
}
private static boolean isAnagram(String a, String b) {
char[] ca = a.toCharArray() ;
char[] cb = b.toCharArray();
Arrays.sort( ca );
Arrays.sort( cb );
return Arrays.equals(ca , cb );
}
}
Метод 4:
public class AnagramsOrNot {
public static void main(String[] args) {
String a = "Protijayi";
String b = "jayiProti";
isAnagram(a, b);
}
private static void isAnagram(String a, String b) {
Map<Integer, Integer> map = new LinkedHashMap<>();
a.codePoints().forEach(code -> map.put(code, map.getOrDefault(code, 0) + 1));
System.out.println(map);
b.codePoints().forEach(code -> map.put(code, map.getOrDefault(code, 0) - 1));
System.out.println(map);
if (map.values().contains(0)) {
System.out.println("Anagrams");
} else {
System.out.println("Not Anagrams");
}
}
}
В Python:
def areAnagram(a, b):
if len(a) != len(b): return False
count1 = [0] * 256
count2 = [0] * 256
for i in a:count1[ord(i)] += 1
for i in b:count2[ord(i)] += 1
for i in range(256):
if(count1[i] != count2[i]):return False
return True
str1 = "Giniiii"
str2 = "Protijayi"
print(areAnagram(str1, str2))
Давайте рассмотрим еще один известный вопрос для интервью: сгруппируйте анаграммы по заданной строке:
public class GroupAnagrams {
public static void main(String[] args) {
String a = "Gini Gina Protijayi iGin aGin jayiProti Soudipta";
Map<String, List<String>> map = Arrays.stream(a.split(" ")).collect(Collectors.groupingBy(GroupAnagrams::sortedString));
System.out.println("MAP => " + map);
map.forEach((k,v) -> System.out.println(k +" and the anagrams are =>" + v ));
/*
Look at the Map output:
MAP => {Giin=[Gini, iGin], Paiijorty=[Protijayi, jayiProti], Sadioptu=[Soudipta], Gain=[Gina, aGin]}
As we can see, there are multiple Lists. Hence, we have to use a flatMap(List::stream)
Now, Look at the output:
Paiijorty and the anagrams are =>[Protijayi, jayiProti]
Now, look at this output:
Sadioptu and the anagrams are =>[Soudipta]
List contains only word. No anagrams.
That means we have to work with map.values(). List contains all the anagrams.
*/
String stringFromMapHavingListofLists = map.values().stream().flatMap(List::stream).collect(Collectors.joining(" "));
System.out.println(stringFromMapHavingListofLists);
}
public static String sortedString(String a) {
String sortedString = a.chars().sorted()
.collect(StringBuilder::new, StringBuilder::appendCodePoint, StringBuilder::append).toString();
return sortedString;
}
/*
* The output : Gini iGin Protijayi jayiProti Soudipta Gina aGin
* All the anagrams are side by side.
*/
}