Question

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

Given the head of a singly linked list, return true if it is a palindrome or false otherwise.

Example 1:

Input: head = [1,2,2,1]
Output: true

Example 2:

Input: head = [1,2]
Output: false

Constraints:

The number of nodes in the list is in the range [1, 10⁵].
0 <= Node.val <= 9

Follow up: Could you do it in O(n) time and O(1) space?

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

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;
    }
}

############

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        ListNode slow = head;
        ListNode fast = head.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode cur = slow.next;
        slow.next = null;
        ListNode pre = null;
        while (cur != null) {
            ListNode t = cur.next;
            cur.next = pre;
            pre = cur;
            cur = t;
        }
        while (pre != null) {
            if (pre.val != head.val) {
                return false;
            }
            pre = pre.next;
            head = head.next;
        }
        return true;
    }
}

// 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:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def isPalindrome(self, head: Optional[ListNode]) -> bool:
        slow, fast = head, head.next
        while fast and fast.next:
            slow, fast = slow.next, fast.next.next
        pre, cur = None, slow.next
        while cur:
            t = cur.next
            cur.next = pre
            pre, cur = cur, t
        while pre:
            if pre.val != head.val:
                return False
            pre, head = pre.next, head.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

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func isPalindrome(head *ListNode) bool {
	slow, fast := head, head.Next
	for fast != nil && fast.Next != nil {
		slow, fast = slow.Next, fast.Next.Next
	}
	var pre *ListNode
	cur := slow.Next
	for cur != nil {
		t := cur.Next
		cur.Next = pre
		pre = cur
		cur = t
	}
	for pre != nil {
		if pre.Val != head.Val {
			return false
		}
		pre, head = pre.Next, head.Next
	}
	return true
}

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function isPalindrome(head: ListNode | null): boolean {
    let slow: ListNode = head,
        fast: ListNode = head.next;
    while (fast != null && fast.next != null) {
        slow = slow.next;
        fast = fast.next.next;
    }
    let cur: ListNode = slow.next;
    slow.next = null;
    let prev: ListNode = null;
    while (cur != null) {
        let t: ListNode = cur.next;
        cur.next = prev;
        prev = cur;
        cur = t;
    }
    while (prev != null) {
        if (prev.val != head.val) return false;
        prev = prev.next;
        head = head.next;
    }
    return true;
}

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {boolean}
 */
var isPalindrome = function (head) {
    let slow = head;
    let fast = head.next;
    while (fast && fast.next) {
        slow = slow.next;
        fast = fast.next.next;
    }
    let cur = slow.next;
    slow.next = null;
    let pre = null;
    while (cur) {
        let t = cur.next;
        cur.next = pre;
        pre = cur;
        cur = t;
    }
    while (pre) {
        if (pre.val !== head.val) {
            return false;
        }
        pre = pre.next;
        head = head.next;
    }
    return true;
};

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution {
    public bool IsPalindrome(ListNode head) {
        ListNode slow = head;
        ListNode fast = head.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode cur = slow.next;
        slow.next = null;
        ListNode pre = null;
        while (cur != null) {
            ListNode t = cur.next;
            cur.next = pre;
            pre = cur;
            cur = t;
        }
        while (pre != null) {
            if (pre.val != head.val) {
                return false;
            }
            pre = pre.next;
            head = head.next;
        }
        return true;
    }
}

234 - Palindrome Linked List

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