for loop

# Check for Palindrome Number with for loop

In this example we shall show you how to check if a palindrome number exists in an array, using a for loop. A palindrome number is a number that is equal to its reverse number. To check if a palindrome number exists in an array, using a for loop one should perform the following steps:

• Create an array of the numbers to be checked. The numbers of the example are int numbers.
• Loop through the array of the numbers to check if a palindrome number exists.
• For every number in the array, reverse it and check if it is equal to its reverse number,

as described in the code snippet below.

```package com.javacodegeeks.snippets.basics;

public static void main(String[] args) {

// numbers to check
int numbers[] = new int[]{ 252, 54, 99, 1233, 66, 9876 };

// loop through the given numbers
for (int i = 0; i < numbers.length; i++) {

int numberToCheck = numbers[i];
int numberInReverse = 0;
int temp = 0;

// a number is a palindrome if the number is equal to it's reversed number

// reverse the number
while (numberToCheck > 0) {
temp = numberToCheck % 10;
numberToCheck = numberToCheck / 10;
numberInReverse = numberInReverse * 10 + temp;
}

if (numbers[i] == numberInReverse) {
System.out.println(numbers[i] + " is a palindrome");
}
else {
System.out.println(numbers[i] + " is NOT a palindrome");
}

}

}

}
```

Output:

``````252 is a palindrome
54 is NOT a palindrome
99 is a palindrome
1233 is NOT a palindrome
66 is a palindrome
9876 is NOT a palindrome```
```

This was an example of how to check if a palindrome number exists in an array, using a for loop in Java.

### Byron Kiourtzoglou

Byron is a master software engineer working in the IT and Telecom domains. He is an applications developer in a wide variety of applications/services. He is currently acting as the team leader and technical architect for a proprietary service creation and integration platform for both the IT and Telecom industries in addition to a in-house big data real-time analytics solution. He is always fascinated by SOA, middleware services and mobile development. Byron is co-founder and Executive Editor at Java Code Geeks.
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