剑指offer 16. 反转链表

class ListNode {
int data;
ListNode next;
}

// 题目：链表翻转
// 解法：设立三个标志直接进行翻转
public class Main {

public static void main(String[] args) {
ListNode n1 = new ListNode();
ListNode n2 = new ListNode();
ListNode n3 = new ListNode();
ListNode n4 = new ListNode();
ListNode t1 = new ListNode();
ListNode t2 = new ListNode();
ListNode t3 = new ListNode();
ListNode t4 = new ListNode();
n1.data = 1;
n2.data = 3;
n3.data = 5;
n4.data = 7;
t1.data = 2;
t2.data = 4;
t3.data = 6;
t4.data = 8;
n1.next = n2;
n2.next = n3;
n3.next = n4;
n4.next = t1;
t1.next = t2;
t2.next = t3;
t3.next = t4;
ListNode result = reverseList(n1);
while(result!=null){
System.out.println(result.data);
result = result.next;
}
}

public static ListNode reverseList(ListNode node) {
if(node == null || node.next == null){
return node;
}
ListNode pre = null;							//前序结点先设置为空
ListNode point = node;							//操作结点为头节点
ListNode pos = point.next;						//pos为当前节点下一个节点
while(pos!=null){								//当pos结点为空则跳出循环
point.next = pre;
pre = point;
point = pos;
pos = point.next;
}
point.next = pre;								//使当前节点与前序结点相连
return point;									//并返回当前节点
}

}