原题出处：https://leetcode.cn/leetbook/read/top-interview-questions-easy/xnnhm6/

解法一：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        Stack<ListNode> stack = new Stack();
        while(head != null) {
            stack.push(head);
            head = head.next;
        }
        if (stack.isEmpty()) {
           return null; 
        }
        ListNode node = stack.pop();
        ListNode result = node;
        while (!stack.isEmpty()) {
            ListNode tmpNode = stack.pop();
            node.next = tmpNode;
            node = node.next;
        }
        node.next = null;
        return result;

    }
}

思路：通过栈解决，“先进后出”，先入栈后出栈，链表就反转过来了。

解法二：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode result = null;
        while(head != null) {
            ListNode tmpNode = head.next;
            head.next = result;
            result = head;
            head = tmpNode;
        }
        return result;
    }
}

思路：使用两个链表，将原来的链表中的节点一个个拆出来，然后拼接到新的链表上即可。

解法三：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode result = reverseList(head.next);
        head.next.next = head;
        head.next = null;
        return result;
    }
}

思路：递归，从链表的最后一个节点重新创建一个链表