LeetCode(回文链表)

题目：

请判断一个链表是否为回文链表。

示例 1:

输入: 1->2
输出: false
示例 2:

输入: 1->2->2->1
输出: true

代码1

//获取链表的长度
    int length = 0;
    ListNode node = head;
    if(null == head.next){
        length = 1;
    }else{
        do{
            length++;
        }while(null != (node = node.next));
    }

    Stack stack = new Stack<>();
    //如果ListNode的长度为单数
    if(1 == length % 2){
        //入栈
        for(int i = 0, j = length / 2; i < j; i++){
            stack.push(head.val);
            head = head.next;
        }

        head = head.next;

        //出栈
        for(int i = length / 2 + 1, j = length; i < j; i++){
            if(head.val != stack.pop()){
                return false;
            }
            head = head.next;
        }
    //如果ListNode的长度为双数
    }else{
        //入栈
        for(int i = 0, j = length / 2; i < j; i++){
            stack.push(head.val);
            head = head.next;
        }
        //出栈
        for(int i = length / 2 + 1, j = length; i <= j; i++){
            if(head.val != stack.pop()){
                return false;
            }
            head = head.next;
        }
    }

    return true;
}

代码2

class Solution {
    public boolean isPalindrome(ListNode head) {
        if (head == null || head.next == null) return true; // 只有一个节点肯定是true
        ListNode fast = head;
        ListNode slow = head;
        ListNode temp = null;
        while (fast != null){
            /**
             * 1.快速找到中间的节点
             * 2.将慢指针指向的前半部分翻转
             * 3.从中间开始向左右两边遍历比较
             */
            if (fast.next != null){ // 偶数个节点
                slow = slow.next;
                fast = fast.next.next;
                // 遍历时顺便翻转前半部分
                head.next = temp; // 翻转代码
                temp = head;
                head = slow;
            } else { // 奇数个节点
                
                slow = slow.next;
                // 遍历时顺便翻转前半部分
                head.next = temp; // 翻转代码
                head = slow;
                break;
            }

        }
        while (temp != null){
            if (temp.val != head.val) return false;
            temp = temp.next;
            head = head.next;
        }
        return true;
    }
}