Question

Formatted question description: https://leetcode.ca/all/234.html

 234	Palindrome Linked List

 Given a singly linked list, determine if it is a palindrome.

 Example 1:

 Input: 1->2
 Output: false

 Example 2:

 Input: 1->2->2->1
 Output: true
 Follow up:
 Could you do it in O(n) time and O(1) space?

 @tag-linkedlist

Algorithm

The principle of using fast and slow pointers to find the midpoint,each time the fast pointer moves two steps and the slow pointer moves one step. When the fast pointer finishes, the position of the slow pointer is the midpoint.

After finding the midpoint, flip the linked list in the second half to compare in the order of palindrome.

Code

Java

public class Palindrome_Linked_List {

    /**
     * Definition for singly-linked list.
     * public class ListNode {
     *     int val;
     *     ListNode next;
     *     ListNode(int x) { val = x; }
     * }
     */
    public boolean isPalindrome(ListNode head) {

        if (head == null) {
            return false;
        }

        // 2 pointers to find mid
        ListNode slowPrev = new ListNode(0);
        ListNode slow = head;
        ListNode fast = head;

        while (fast != null && fast.next != null) {
            fast = fast.next.next;

            slowPrev = slow;
            slow = slow.next;
        }

        // now slow aat middle
        slowPrev.next = null; // cut

        ListNode reversed2ndHalf = reverseNodeList(slow);
        ListNode p1 = head;
        ListNode p2 = reversed2ndHalf;

        while (p1 != null && p2 != null) {
            if (p1.val != p2.val) {
                return false;
            }

            p1 = p1.next;
            p2 = p2.next;
        }

        return true;
    }

    private ListNode reverseNodeList(ListNode head) {

        ListNode dummy = new ListNode(0);

        ListNode current = head;

        while (current != null) {
            ListNode currenHead = dummy.next;
            ListNode nextHead = current.next;

            current.next = null; // cut current node out

            dummy.next = current;
            current.next = currenHead;
            current = nextHead;
        }

        return dummy.next;
    }
}

// OJ: https://leetcode.com/problems/palindrome-linked-list/
// Time: O(N)
// Space: O(1)
class Solution {
private:
    int getLength(ListNode *head) {
        int len = 0;
        for (; head; head = head->next, ++len);
        return len;
    }
    ListNode *reverse(ListNode *head) {
        ListNode dummy(0);
        while (head) {
            ListNode *node = head;
            head = head->next;
            node->next = dummy.next;
            dummy.next = node;
        }
        return dummy.next;
    }
public:
    bool isPalindrome(ListNode* head) {
        if (!head) return true;
        int len = (getLength(head) + 1) / 2;
        ListNode *p = head, *q;
        while (--len > 0) p = p->next;
        q = p->next;
        p->next = NULL;
        q = reverse(q);
        while (head && q) {
            if (head->val != q->val) return false;
            head = head->next;
            q = q->next;
        }
        return true;
    }
};

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
  def isPalindrome(self, head):
    """
    :type head: ListNode
    :rtype: bool
    """

    def reverseList(root):
      pre = None
      cur = root
      while cur:
        tmp = cur.next
        cur.next = pre
        pre = cur
        cur = tmp
      return pre

    slow = fast = head
    while fast and fast.next:
      slow = slow.next
      fast = fast.next.next

    newHead = reverseList(slow)
    p1 = head
    p2 = newHead
    while p1 and p2:
      if p1.val != p2.val:
        return False
      p1 = p1.next
      p2 = p2.next
    return True

234 - Palindrome Linked List

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