Determine if the linked list is a palindrome

Palindrome of linked list

The palindrome structure of the linked list—Niuke.com link

Topic content: 对于一个链表，请设计一个时间复杂度为O(n),额外空间复杂度为O(1)的算法，判断其是否为回文结构。 给定一个链表的头指针A，请返回一个bool值，代表其是否为回文结构。保证链表长度小于等于900。
Input and output example:

输入：
1->2->2->1
输出：
true

1. Code implementation

Method 1: First reverse the linked list, and then traverse and compare whether they are equal;

Reverse linked list blog, if you are interested, you can check it out

public boolean chkPalindrome(ListNode A) {
    
    
        // write code here
        //1.思路一：首先将链表反转，然后比较是否相等
        ListNode B = reverseList(A);//将链表进行反转

        //开始遍历，比较俩个链表
        while (A != null){
    
    //循环条件也可以是B不等于null，因为反转过，链表长度肯定一样
            if (A.val != B.val){
    
    
                return false;
            }else {
    
    
                A = A.next;
                B = B.next;
            }
        }
        return true;
    }
    
    //反转链表的方法
    public ListNode reverseList(ListNode head) {
    
    
        //迭代的方法需要保存前一个节点
        ListNode pre = null;
        ListNode cur = head;//用来遍历的节点

        while (cur != null){
    
    
            //1.首先将cur后面的链表保存下来
            ListNode next = cur.next;
            //2.然后将cur.next变为它的前一个节点
            cur.next = pre;
            //3.然后pre保存新链表(也就是反转链表的头节点)
            pre = cur;
            //4.cur继续遍历原来的链表
            cur = next;
        }
        //5.遍历完成，返回新链表的头节点
        return pre;
    }

Method 2: The first traversal puts the value of the node into the stack, and then compares it with the elements in the stack after traversing once (using the first-in-last-out of the stack)

public boolean chkPalindrome(ListNode A) {
    
    
        // write code here
        Stack<Integer> stack = new Stack<>();
        //1.第一遍遍历，入栈
        ListNode cur = A;
        while (cur != null){
    
    
            stack.push(cur.val);
            cur = cur.next;
        }
        
        //2.第二遍遍历，进行比较
        while (A != null){
    
    
            if (A.val != stack.pop()){
    
    
                return false;
            }
            A = A.next;
        }
        return true;
    }