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recursivefunction.cpp
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recursivefunction.cpp
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/*********************************************************************
* ASSIGNMENT 07 - RECURSIVE FUNCTION
*
* This program demonstrates the use of a recursive function to check whether
* a string entered by the user is a palindrome or not.
*
* It includes two functions:
* (1) RemoveWhitespace: As it's name implies, this function removes whitespace
* from the user entered string as well as punctuation and returns a
* filtered string that is all lowercase.
* (2) isPalindrome: This function checks whether the filtered string is a
* palindrome or not.
*
* Written by Sindhura L Boppudi (slb170330)
* CS 1337. 004
* Professor: John Cole
*********************************************************************/
#include "stdafx.h"
#include <iostream>
#include <iomanip>
#include <string>
#include <cstring>
#include <cctype>
using namespace std;
//FUNCTION PROTOTYPE
void removeWhitespace(string&);
bool isPalindrome(string);
int main()
{
string userInp;
while (true)
{
//Ask user for string
cout << "Enter a string: ";
getline(cin, userInp);
//If user enters string of length 0, exit program all together
if (userInp == "")
{
return 0;
}
//Pass user input by reference to function that removes whitespace and punctuation
removeWhitespace(userInp);
//Call the recursive function
if (isPalindrome(userInp))
{
cout << "The string you have entered is a palindrome." << endl;
}
else
cout << "The string you have entered is not a palindrome."<< endl;
cout << "" << endl;
}
return 0;
}
void removeWhitespace(string &s)
{
//Create loop to go through each character of the string.
//If the character is a whitespace or puntuation, shift all the other characters one space to the left and delete the last character
//Start checking again from the beggining
for (int ix = 0; ix < s.length(); ix++)
{
if (isspace(s[ix]) || ispunct(s[ix]))
{
for (int iy = ix; (iy+1) < s.length(); iy++)
{
s[iy] = s[iy + 1];
}
s.erase((s.length()-1), 1);
ix = 0; //Check from the beginning again
}
}
//Convert all characters to lower case so they can be compared directly later in the program
for (int ix = 0; ix < s.length(); ix++)
{
s.at(ix) = tolower(s.at(ix));
}
}
bool isPalindrome(string s)
{
int numChars = s.length();
if (s.at(0) == s.at(numChars - 1)) //Address the fact that numchars will be one more than the array position of char in string.
{
//When string has odd number of characters, it will be reduced to one character if it is a palindrome
if (numChars <= 1)
{
return true;
}
//When there are only two characters in the string, if they are equal then it is a palindrome
else if (numChars == 2)
{
return true;
}
//This will create a substring with all the same characters of string s without the characters at the two ends that have already been established to be equal to each other
//Example: "kayak" will become "aya" and isPalindrome function will be called for "aya" string.
s = s.substr(1, numChars - 2);
return isPalindrome(s);
}
else
return false;
}