Palindrome Linked List

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

Follow up:

Could you do it inO(n)time andO(1)space?

Tips:

找中点，reverse前一半，比较。需要注意fast所指的对象，在一个循环内完成的话需要prev。

Code：

public class Solution {
    public boolean isPalindrome(ListNode head) {
        if (head == null) return true;
        ListNode slow = head;
        ListNode fast = head;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        int count = 0;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            count++;
        }
        while (count > 1) {
            ListNode temp = head.next;
            head.next = head.next.next;
            temp.next = dummy.next;
            dummy.next = temp;
            count--;
        }
        head = dummy.next;
        if (fast != null) slow = slow.next;
        while (slow != null) {
            if (head.val != slow.val) return false;
            head = head.next;
            slow = slow.next;
        }
        return true;
    }
}

别人的，一个循环：

public class Solution {
    public boolean isPalindrome(ListNode head) {
        if(head == null) {
            return true;
        }
        ListNode p1 = head;
        ListNode p2 = head;
        ListNode p3 = p1.next;
        ListNode pre = p1;
        //find mid pointer, and reverse head half part
        while(p2.next != null && p2.next.next != null) {
            p2 = p2.next.next;
            pre = p1;
            p1 = p3;
            p3 = p3.next;
            p1.next = pre;
        }

        //odd number of elements, need left move p1 one step
        if(p2.next == null) {
            p1 = p1.next;
        }
        else {   //even number of elements, do nothing

        }
        //compare from mid to head/tail
        while(p3 != null) {
            if(p1.val != p3.val) {
                return false;
            }
            p1 = p1.next;
            p3 = p3.next;
        }
        return true;

    }
}