题解 | #反转链表#

大致思路

work指针指向当前即将成为头结点的指针，然后让p指向该节点的next(注意，该节点只是即将成为头结点，而不是已经成为头结点)
当work插到头了之后，让head指向该节点（即head一直指向新链表的表头），此时p指向着原链表的剩余链表的头位置，然后又让work指向p，p指向work的next，继续处理work，循环直至work为空，最后返回head。

/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
public class Solution {
    public ListNode ReverseList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode work = head.next;
        head.next = null;
        ListNode p;
        while (work != null) {
            p = work.next;
            work.next = head;
            head = work;
            work = p;
            if (work != null) {
                p = work.next;
            }
        }
        return head;
    }
}