Dada una string, la tarea es contar todas las substrings de palíndromo en una string dada.
Input : aba Output : 4 Explanation : All palindrome substring are : "aba" , "a" , "b", "a" Input : TENET Output : 7 Explanation : All palindrome sub-string are : "T" , "E" , "N", "E", "T" , "ENE" , "TENET"
Acercarse:
- Tome las substrings de la string dada.
- Ahora compruebe si la substring es Palindrome o no.
- Si la substring es Palindrome, incremente el conteo en 1; de lo contrario, no se incrementa el conteo.
- Devuelve el recuento, ya que representa el número de substrings.
- Imprima y visualice en la consola.
Ejemplo:
Java
// Java Program to Find all palindromic sub-strings
// of a given string
// Importing input output classes
import java.io.*;
// Main class
// To check for pallindromic sub-strings
public class GFG {
// Method 1
// To check for substring
public static boolean check(String subS)
{
int size = subS.length();
// Iterating over string till midway to
// check pallindromic behavior
for (int i = 0; i < size / 2; i++) {
if (subS.charAt(i)
!= subS.charAt(size - i - 1)) {
// Non-pallindromic behavior
return false;
}
}
// Pallindromic behavior
return true;
}
// Method 2
// Main driver method
public static void main(String[] args)
{
// Custom input string
String str = "MALAYALAM";
int count = 0;
// Outer loop iterating over input string
for (int i = 0; i < str.length(); i++) {
// Inner loop iterating from current starting
// character of outer loop current starting
// character
for (int j = i; j < str.length(); j++) {
// Getting the substrings
String subString = str.substring(i, j + 1);
// Checking whether the substring is
// pallindrome
if (check(subString)) {
// Increment the count only if
// substring is pallindrome
count++;
}
}
}
// Print and display the number of substrings that
// are pallindromic
System.out.println(
"No.of palindromic substrings in the given string are "
+ count);
}
}
Producción
No.of palindromic substrings in the given string are 15
Publicación traducida automáticamente
Artículo escrito por ganiprabhakar y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA