判断一个链表是否是回文链表，时间复杂度O(n)，空间复杂度O(1)

leetcode234

思路：（1）快慢指针找到链表中点（2）原地反转中点之后的链表（3）比较两段链表的值（4）反转的链表再反转回去

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
 // O(n) 时间复杂度和 O(1) 空间复杂度
class Solution {
    public boolean isPalindrome(ListNode head) {
        if(head == null){
            return true;
        }
        //1.快慢指针找到链表中点
        ListNode slow = head;
        ListNode fast = head;
        while(fast.next != null && fast.next.next != null){
            slow = slow.next;
            fast = fast.next.next;
        }
        //2.将后半部分的链表反转
        ListNode curNode = slow.next;
        ListNode pre = null;
        while(curNode != null){
            ListNode nextNode = curNode.next;
            curNode.next = pre;
            pre = curNode;
            curNode = nextNode;
        }
        slow.next = pre;
        ListNode leftNode = head;
        ListNode rightNode = pre;
        while(leftNode != null && rightNode != null){
            if(leftNode.val != rightNode.val){
                return false;
            }
            leftNode = leftNode.next;
            rightNode = rightNode.next;
        }
        //3.将反转后的链表反转回来
        curNode = slow.next;
        pre = null;
        while(curNode != null){
            ListNode nextNode = curNode.next;
            curNode.next = pre;
            pre = curNode;
            curNode = nextNode;
        }
        slow.next = pre;
        return true;
    }
}