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ashutosh049
ashutosh049

Posted on • Edited on

check if Linked List is palindrome or not

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Approach

  1. Find the middle Node
  2. Reverse the latter half of list
  3. Compare each node of the original list to reversed list.
    • Note: the list returned after reversing, breaks the original list into 2 parts, we return the prev which is reversed

Node Def

 class ListNode {
      public int val;
      public ListNode next;
      ListNode(int x) { val = x; next = null; }
  }

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Runner class

public class LinkedListPalindrome{
    public int isPalindrome(ListNode root) {

        // If no Node, false always
        if (root == null)
            return 0;

        // If single Node only, true always
        if (root.next == null)
            return 1;

        ListNode curr = root;

        //1. Find the middle Node
        ListNode midNode = findMidNode(curr);
        //2. Reverse the latter half of list
        ListNode revNode = reverseNodeList(midNode);
        //3. Compare each node 
        curr = root;
        while(curr != null && revNode != null){
            if(curr.val != revNode.val){
                return 0;
            }
            curr = curr.next;
            revNode = revNode.next;
        }
        return 1;
    }
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find the middle node

  1. If list size is even , middle Node will be len/2 th Node
  2. If list size is odd, middle Node will be len/2 th Node(floor Node)
    /**
     * Use 2 pointer approach
     * fast pointer nad slow pointer
     */
    private ListNode findMidNode(ListNode root){
        ListNode sp = root;
        ListNode fp = root;
        while(fp.next != null && fp.next.next !=null){
            fp = fp.next.next;
            sp = sp.next;
        }
        return sp;
    }
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    private ListNode reverseNodeList(ListNode root){
        ListNode curr = root;
        ListNode prev = null;
        ListNode next = null;

        while(curr != null ){
            next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        return prev;
    }
}
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