题目

isPalindrome

这个题用链表方式可以实现，但效率很低

因为是回文链表，使用快慢指针，每次循环让fast走两步，slow走一步

反转后半段链表，反转起始节点为当前slow节点

题目解答

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public class Solution {
    public boolean isPalindrome(ListNode head) {
        ListNode slow = head;
        ListNode fast = head;

        // fast结束时候slow在中间
        while (slow != null && fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }

        // 反转后半段
        ListNode newHead = reverseList(slow);

        while (newHead != null && head != null) {
            if (newHead.val != head.val) {
                return false;
            }

            newHead = newHead.next;
            head = head.next;
        }
        return true;
    }

    public ListNode reverseList(ListNode head) {
        return reverseList(head, null);
    }

    public ListNode reverseList(ListNode head, ListNode newHead) {

        if (head == null) {
            return newHead;
        }

        ListNode tmp = head.next;

        head.next = newHead;
        newHead = head;

        return reverseList(tmp, newHead);
    }
}