题目

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

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

O(n) time, O(n) space

class Solution {
    public boolean isPalindrome(ListNode head) { return recur(head, head) == null;}
    public ListNode recur(ListNode curr, ListNode head) {
        if(curr == null) return head;
        ListNode ret = recur(curr.next, head);
        return (ret.val == curr.val)? ret.next : ret;
    }
}

O(n) time, O(1) space

class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode next;
        ListNode curr = head;
        ListNode newlist = null;

        while(curr != null) {
            next = curr.next;
            curr.next = newlist;
            newlist = curr;
            curr = next;
        }
        return newlist;
    }
    public boolean isPalindrome(ListNode head) { 
        // Find middle node
        int size = 0, count = 0;
        ListNode curr = head, middle_node = null, curr2 = null;
        while(curr != null) {
            size++;
            curr = curr.next;
        }
        int middle = (size % 2 == 0) ? size / 2 : size / 2 + 1;
        curr = head;
        while(curr != null) {
            if(count == middle) {
                middle_node = curr;
            }
            curr = curr.next;
            count++;
        }
        // Reverse list starting from the middle...
        ListNode reversed = reverseList(middle_node);
        
        // You want reversed == head for the first size / 2 elements
        curr = head;
        curr2 = reversed;
        count = 0;
        while(count < size / 2) {
            if(curr.val != curr2.val) return false;
            curr = curr.next;
            curr2 = curr2.next;
            count++;
        }
        return true;
    }
}