constchar* itss = ss.str().c_str();constchar* ites = itss +strlen(itss) -1;intis_palindrome =1;while(itss <= ites) {if(*itss != *ites) { is_palindrome =0;break; } itss++; ites--; } cout <<"your number is "<< (is_palindrome ?"a palindrome":"NOT a palindrome") <<"...
Let’s take one more example specifically using a while loop that will also explain the algorithm we discussed in the introduction. We will take a number as an input from the user and check if it is a palindrome or not. Example #2 Let us check if a number is a palindrome or not by...
//Java program to check whether a number is a Palindrome or not using While Loop import java.util.*; public class PalindromeNumberExample { public static void main(String[] args) { int r=0, rem, num; Scanner s = new Scanner(System.in); System.out.print("Enter number that has to be...
UPD. Keep in mind however that this is pretty much "cheating" approach, a demonstration of smart usage of language features, but not the most practical algorithm (time O(n), space O(n)). For real life application or coding interview you should definitely use loop solution. The one posted...
In the example, we turn the original string to an array of characters. In a while loop, we start to compare the characters from the both sides of the string; starting with the leftmost and rightmost characters. We go until the middle of the string. ...
Check Palindrome Number using Java program//Java program for Palindrome Number. import java.util.*; public class Palindrome { public static void main(String args[]){ int num,tNum,sum; Scanner bf=new Scanner(System.in); //input an integer number System.out.print("Enter any integer number:...
By which I mean if we take 3 numbers in sequence like 1,2, 3 and create a palindrome by using the ordering a,b,c,b,a and increase that 3 number set infinitely, then add the resulting palindromes, more palindromes will occur but they are sparse. This can't be proved deductively, ...
You can do this using the erase() function in the string class. Loop through the string, and if string[i] is punctuation, number or a space, then erase that character. Return this modified string back to main. (2) Pass this modified string (now with no punctuation or spaces or ...
Here, we are going to learn how to check whether a given number is palindrome or not in JavaScript.
1classSolution {2func isPalindrome(_ s: String) ->Bool {3ifs.count ==0{4returntrue5}67varchas = s.cString(using:.ascii)!89varleft =010varright = s.count -11112varloop =true1314whileloop {15ifleft >right {16loop =false17continue18}1920let leftChar =chas[left]2122if!checkIsNormalCha...