Respuesta :
Answer:
Here is the recursive function:
bool isPalindrome(string& MyString,int start, int end){ // function that takes as argument a string and two arguments that are bounds indices
while(start < MyString.length() && !isalpha(MyString[start])){ //while loop continues to execute as long as start does not exceed the length of the string AND does not point to a alphabetic letter
start++; } // increments start by 1 moving it to point the next character
while(end >= 0 && !isalpha(MyString[end])){ //while loop continues to execute as long as end does not gets less than 0 AND does not point to a alphabetic letter
end--; } // decrements end by 1 moving it to point the next character backwards
if(start > end){ //the base case
return true;
} else { //recursive case
return ( ( toupper(MyString[start]) == toupper(MyString[end]) ) && isPalindrome(MyString,start+1,end-1)); } } //The recursive case is to AND the result of comparing the alphabetic letters at start and end with result of the function (called recursively) after without both of the letters
Explanation:
Here is the complete C++ program:
#include <iostream> //to use input output functions
#include <cctype> //functions to manipulate characters
using namespace std; //to identify objects as cin cout
bool isPalindrome(string&,int start, int end); // function prototype
int main(){ //start of main function
string MyStrings; //declare a string type variable to hold string of characters
cout<<"Enter a string : "; //prompts user to enter a string
getline(cin,MyStrings ); //gets and reads the string from user
while(MyStrings != "quit"){ // continues to take input from user until user enters quit
if(isPalindrome(MyStrings ,0,MyStrings.length()-1)== true){ //calls function that checks if the string is a palindrome
cout<<MyStrings<<" is a Palindrome\n";} //if the input string is a palindrome then display this message
else
{cout<<MyStrings<<" is not a Palindrome\n";} //if the input string is not a palindrome then displays this message
cout<<"Enter a string : "; //prompts user to enter a string
getline(cin,MyStrings); } } //reads a string from user
bool isPalindrome(string& MyString,int start, int end){ //function definition
while(start < MyString.length() && !isalpha(MyString[start])){
start++; }
while(end >= 0 && !isalpha(MyString[end])){
end--; }
if(start > end){
return true;
} else {
return ( ( toupper(MyString[start]) == toupper(MyString[end]) ) && isPalindrome(MyString,start+1,end-1)); } }
I will explain the working of the function:
First the start variable is pointed to the start of the MyString and the end variable points to the last character of MyString
while(start < MyString.length() && !isalpha(MyString[start]))
The above statement has a while loop which continues to execute as long as start does not exceed the string length and it does not point to a letter. It has a AND logical operator that means both the above condition have to be true in order to continue this while loop. Here isalpha() method is used that checks if the letter/character of MyString is an alphabet or not. At each iteration start is incremented to 1 to point the next character of MyString.
while(end >= 0 && !isalpha(MyString[end]))
The above statement has a while loop which continues to execute as long as end does not becomes less than 0 and it does not point to a letter, Here isalpha() method is used that checks if the letter/character of MyString is an alphabet or not. At each iteration end is decremented to 1 to point the next character of MyString in backward or in reverse order.
if(start > end){
The above statement is a base case which is the IF condition that checks if the start is greater than end. This basically shows that there are no more alphabetic characters and this base case returns true as it reads the same from both sides.
else {
return ( ( toupper(MyString[start]) == toupper(MyString[end]) ) && isPalindrome(MyString,start+1,end-1)); } }
The above statement is a recursive part. It uses AND logical operator and compares the characters at start and end with the result of the function after excluding both of the characters pointed by start and end.
The program along with its output is attached.
