java算法题-反转链表

直接上代码，因为是在线编辑器，很多package都省略了，主要是看关键代码；通过递归方式反转链表

 

```java

public class HelloWorld{

public static void main(String []args){
ListNode node5 = new ListNode(5,null);
ListNode node4 = new ListNode(4,node5);
ListNode node3 = new ListNode(3,node4);
ListNode node2 = new ListNode(2,node3);
ListNode node1 = new ListNode(1,node2);

printListNode(node1);

ListNode res = reverseList(node1);
System.out.println();
System.out.print("reverser result is:");
printListNode(res);
}

private static void printListNode(ListNode node){
while(node!=null){
System.out.print(node.val +(node.next!=null ? "->":""));
node = node.next;
}
}

public static class ListNode{
public int val;
public ListNode next;
public ListNode(int val,ListNode next){
this.val = val;
this.next = next;
}
}

private static ListNode reverseList(ListNode head){

if(head null || head.nextnull) return head;

ListNode listRes = reverseList(head.next);

head.next.next = head;
head.next = null;

return listRes;
}
}

补充：迭代法

```java

    public static ListNode diedai(ListNode head){
        ListNode preNode=null,next,curNode;        
        curNode = head;
        while(curNode!=null){
            next = curNode.next;
            curNode.next = preNode;
            preNode = curNode;
            curNode = next;
        }
        return preNode;
    }
```java

### 运行结果：
1->2->3->4->5
reverser result is:5->4->3->2->1