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?

Analysis

由于单链表没有反向指针，因此就需要将链表分为左右两部分进行比较。找中点可以用快满指针，找到之后翻转右边一半的链表，这时就可以进行一对一比较了。

参考： https://leetcode.com/problems/palindrome-linked-list/discuss/64501/Java-easy-to-understand

In the beginning, set two pointersfastandslowstarting at the head.

1 -> 1 -> 2 -> 1 -> null 
sf

(1) Move:fastpointer goes to the end, andslowgoes to the middle.

1 -> 1 -> 2 -> 1 -> null 
          s          f

(2) Reverse: the right half is reversed, andslowpointer becomes the 2nd head.

1 -> 1    null <- 2 <- 1           
h                      s

(3) Compare: run the two pointersheadandslowtogether and compare.

1 -> 1    null <- 2 <- 1             
     h            s

Solution

Find Mid Point + Reverse + Compare --- O(n) time, O(1) space

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        ListNode fast, slow;
        fast = head;
        slow = head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        if (fast != null) {
            slow = slow.next;
        }
        slow = reverse(slow);
        fast = head;

        while (fast != null && slow != null) {
            if (fast.val != slow.val) {
                return false;
            }
            fast = fast.next;
            slow = slow.next;
        }
        return true;
    }

    ListNode reverse(ListNode head) {
        ListNode prev = null;
        ListNode curr = head;

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

Don't need to make the right half smaller: https://leetcode.com/problems/palindrome-linked-list/discuss/64501/Java-easy-to-understand/66206

public boolean isPalindrome(ListNode head) {
        if (head == null || head.next == null) {
            return true;
        }
        ListNode fast = head.next;
        ListNode slow = head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        slow = reverse(slow);
        while (head != null && slow != null) {
            if (head.val != slow.val) {
                return false;
            }
            head = head.next;
            slow = slow.next;
        }
        return true;
    }

Reference

https://leetcode.com/problems/palindrome-linked-list/discuss/64501/Java-easy-to-understand