Thursday, July 12, 2012

Write a program to check if a linked list is a palindrome

The approach:-
  1. Solution for this problem is recursion with a reference to head node of the list.

  2. Whenever the recursive function returns compare the node values and move the reference to the next node.

C++ program to check if a linked list is a palindrome

#include <iostream>
using namespace std;

// General list node
class Node {
    public:
        Node() { mData = -1; mNext = NULL; }
        ~Node() {}
        int data() { return mData; }
        void setData(int data) { mData = data; }
        Node* next() { return mNext; }
        void setNext(Node* next) { mNext = next; }
    private:
        int mData;
        Node* mNext;
};

// General list class
class List {
    public:
        List() { mHead = NULL; mCount = 0; }
        ~List() {}
        Node* head() { return mHead; }
        void setHead(Node* head) { mHead = head; }

        void append(int data) {
            Node* tmp = new Node();
            tmp->setData(data);
            if ( mHead == NULL )
                    mHead = tmp;
            else {
                Node* ptr = mHead;
                while ( ptr->next() != NULL ) {
                        ptr = ptr->next();
                }
                ptr->setNext(tmp);
            }
        }

        void prepend(int data) {
            Node* tmp = new Node();
            tmp->setData(data);
            tmp->setNext(mHead);
            mHead = tmp;
        }

        void print() {
            Node* ptr = mHead;
            while ( ptr != NULL ) {
                cout << ptr->data() << " ";
                ptr = ptr->next();
            }
            cout << endl;
        }

        bool isPalindrome(Node*& n1, Node* n2) {
            if ( ! n2 )
                return true;

            if ( ! isPalindrome(n1, n2->next()) )
                return false;

            bool flag = ( n1->data() == n2->data() );

            n1 = n1->next();

            return flag;
        }

    private:
        Node* mHead;
        int mCount;
};

int main() {
    List* l = new List();
    l->append(100);
    l->append(200);
    l->append(300);
    l->append(400);
    l->append(300);
    l->append(200);
    l->append(100);
    l->print();

    Node* n1 = l->head();
    bool stat = l->isPalindrome(n1, n1);
    if ( stat ) {
        cout << "Linked list is palindrome" << endl;
    }
    else {
        cout << "Linked list is not a palindrome" << endl;
    }

    delete l;
}

Output:-
100 200 300 400 300 200 100
Linked list is palindrome

4 comments :

  1. Is,
    delete l;
    Freeing all nodes in linked list ? If yes, please explain how ?

    ReplyDelete
  2. can someone explain the code? how n2 is traversing in reverse?

    ReplyDelete
  3. here there is a problem.
    Example: madam
    m==m. a===a, d==d then what is the use of checking again a==a ,m==m this check are of no use.
    Can we avoid this extra check.
    Thanks

    ReplyDelete
  4. This comment has been removed by the author.

    ReplyDelete

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