LeetCode 234. Palindrome Linked List

原题链接在这里：https://leetcode.com/problems/palindrome-linked-list/

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?

题解：

原题要求time O(n), space O(1). 所以不能用额外空间。

先找到中点, reverse中点后面的list部分，再与head开始逐个比较val. 其中reverse部分可以参见Reverse Linked List.

Time Complexity: O(n). Space O(1).

AC Java:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        if(head == null || head.next == null){
            return true;
        }
        
        ListNode mid = findMid(head);
        ListNode midNext = mid.next;
        mid.next = null;
        
        midNext = reverse(midNext);
        
        while(head != null && midNext != null){
            if(head.val != midNext.val){
                return false;
            }
            
            head = head.next;
            midNext = midNext.next;
        }
        
        return true;
    }
    
    private ListNode findMid(ListNode head){
        if(head == null || head.next == null){
            return head;
        }
        
        ListNode runner = head;
        ListNode walker = head;
        while(runner.next != null && runner.next.next != null){
            walker = walker.next;
            runner = runner.next.next;
        }
        
        return walker;
    }
    
    private ListNode reverse(ListNode head){
        if(head == null || head.next == null){
            return head;
        }
        
        ListNode tail = head;
        ListNode cur = tail;
        ListNode pre;
        ListNode temp;
        while(tail.next != null){
            pre = cur;
            cur = tail.next;
            temp = cur.next;
            cur.next = pre;
            tail.next = temp;
        }
        
        return cur;
    }
}