LeetCode——回文链表

Q：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

A：
1.reverse以后对比。
因为这里是递归reverse，所以之前要先复制一个原链表。

    public boolean isPalindrome(ListNode head) {
        if (head == null || head.next == null)
            return true;
        ListNode node = head.next;
        ListNode head1 = new ListNode(head.val);
        ListNode temp = head1;
        while(node!=null){
            int val = node.val;
            temp.next = new ListNode(val);
            temp = temp.next;
            node = node.next;
        }
        ListNode head2 = reverse(head);
        ListNode node1 = head1;
        ListNode node2 = head2;
        while (node1 != null) {
            if (node1.val != node2.val)
                return false;
            node1 = node1.next;
            node2 = node2.next;
        }
        return true;
    }

    private ListNode reverse(ListNode head) {
        if (head == null || head.next == null)
            return head;
        ListNode node = reverse(head.next);
        head.next.next = head;
        head.next = null;
        return node;
    }

2.双指针
主要是找到中点再reverse，我觉得这样就没什么意思了……

    private ListNode reverse(ListNode head) {
        if (head == null || head.next == null)
            return head;
        ListNode node = reverse(head.next);
        head.next.next = head;
        head.next = null;
        return node;
    }

    public boolean isPalindrome(ListNode head) {
        if (head == null || head.next == null)
            return true;
        ListNode slow = head;
        ListNode fast = head.next;
        while (fast != null && fast.next != null) {//找到中点，如果是偶数，为前一个点
            slow = slow.next;
            fast = fast.next.next;
        }
        fast = reverse(slow.next);
        slow.next = null;//cut前后两段
        slow = head;
        while(fast!=null){//这里一定要是fast的长度，fast是后半段，slow是前半段，若为奇数长度链表，slow更长
            if(slow.val!=fast.val)
                return false;
            slow = slow.next;
            fast = fast.next;
        }
        return true;
    }