问题：
实现单链表反转
答案：
链表准备
class Node {private int Data;// 数据域private Node Next;// 指针域public Node(int Data) {// super();this.Data = Data;}public int getData() {return Data;}public void setData(int Data) {this.Data = Data;}public Node getNext() {return Next;}public void setNext(Node Next) {this.Next = Next;}
}
public static void main(String[] args) {Node head = new Node(0);Node node1 = new Node(1);Node node2 = new Node(2);Node node3 = new Node(3);head.setNext(node1);node1.setNext(node2);node2.setNext(node3);// 打印反转前的链表Node h = head;while (null != h) {System.out.print(h.getData() + " ");h = h.getNext();}// 调用反转方法head = reverse(head);System.out.println("\n**************************");// 打印反转后的结果while (null != head) {System.out.print(head.getData() + " ");head = head.getNext();}}递归反转法：在反转当前节点之前先反转后续节点。这样从头结点开始，层层深入直到尾结点才开始反转指针域的指向。简单的说就是从尾结点开始，逆向反转各个结点的指针域指向
private static Node reverse1(Node head) {if(head==null||head.getNext()==null){return head;}Node reHead = reverse1(head.getNext());head.getNext().setNext(head);head.setNext(null);return reHead;}
遍历反转法：递归反转法是从后往前逆序反转指针域的指向，而遍历反转法是从前往后反转各个结点的指针域的指向
private static Node reverse2(Node head) {if(head==null){return head;}Node pre = head;Node cur = head.getNext();Node temp;while (cur!=null) {temp = cur.getNext();cur.setNext(pre);pre = cur;cur=temp;}head.setNext(null);return pre;}
