题目：

定义一个函数，输入一个链表的头结点，反转该链表并输出反转后链表的头结点。

分析：

原地逆置链表，需要操作3个链表的关系。假设i->j->k，当把j指向i的时候，就断开了j->k的指针，为了还能找到k，就需要在断开之前，将k保存下来。

解法：

package com.wsy;

class Node {
    private int value;
    private Node next;

    public Node() {
    }

    public Node(int value, Node next) {
        this.value = value;
        this.next = next;
    }

    public int getValue() {
        return value;
    }

    public void setValue(int value) {
        this.value = value;
    }

    public Node getNext() {
        return next;
    }

    public void setNext(Node next) {
        this.next = next;
    }
}

public class Main {
    public static void main(String[] args) {
        Node head = init();
        print(head);
        head = reverse(head);
        print(head);
    }

    public static Node init() {
        Node node6 = new Node(6, null);
        Node node5 = new Node(5, node6);
        Node node4 = new Node(4, node5);
        Node node3 = new Node(3, node4);
        Node node2 = new Node(2, node3);
        Node node1 = new Node(1, node2);
        return node1;
    }

    public static void print(Node node) {
        while (node != null) {
            System.out.print(node.getValue() + "->");
            node = node.getNext();
        }
        System.out.println();
    }

    public static Node reverse(Node node) {
        if (node == null) {
            return null;
        }
        Node prev = null;
        Node current = node;
        while (current != null) {
            Node next = current.getNext();
            current.setNext(prev);
            prev = current;
            current = next;
        }
        return prev;
    }
}