Question

我正在编写用于在给定节点之间反转双向链表的代码。

给出此链接列表1->2->3->4->5，

函数reverse（2,4）应该导致 1->4->3->2->5。

该函数需要两个节点，但不是索引。

这就是我所拥有的。我尝试使用Eclipse调试器来找出无限期运行的原因。出于某些原因，我看到节点已损坏（当我逐步执行下面标记的行时，Eclipse没有显示数据）

public static void reverse(Node head, Node tail){
        Node prev=head.prev;
        Node current=head,next;
        while(current!=tail.next){
            next = current.next;
            current.next=prev;
            current.prev=next; //No data stepping after this point
            prev=current;
            current=next;
        }
    }

public static void main(String... args){
        Node one = new Node(1);
        Node two = new Node(2);
        Node three = new Node(3);
        Node four = new Node(4);
        Node five = new Node(5);

        five.setNodes(four, null);
        four.setNodes(three, five);
        three.setNodes(two,four);
        two.setNodes(one,three);
        one.setNodes(null, two);

        System.out.println("before reversing...");
        System.out.println(one);

        System.out.println("After reversing...");
        reverse(two,four);
        System.out.println(one);

}

这是我的节点类

class Node {
     Node prev;
     Node next;
     int data;

     public Node(Node prev, Node next, int val){
         this.prev=prev;
         this.next=next;
         this.data=val;
     }

     public Node(int val){
         this(null,null,val);
     }

     public void setNodes(Node prev, Node next){
         this.next=next;
         this.prev=prev;
         }

     public String toString(){
         String toReturn = "[ " + this.data;

         Node current=this.next;
         while (current!=null){
             toReturn+=" -> " + current.data;
             current=current.next;
         }
         toReturn+=" ]";
         return toReturn;
     }

}

Answer 1

试试这个：

public static void reverse(Node head, Node tail) {
    Node first = head;
    Node last = tail;
    while (first != last && first.prev != last) {
        Node preFirst = first.prev;
        Node postFirst = first.next;
        Node preLast = last.prev;
        Node postLast = last.next;
        if (preFirst != null) {
            preFirst.next = last;
        }
        if (postLast != null) {
            postLast.prev = first;
        }
        last.prev = preFirst;
        first.next = postLast;
        if (last != postFirst){
            last.next = postFirst;
            postFirst.prev = last;
            first.prev = preLast;
            preLast.next = first;
            first = postFirst;
            last = preLast;
        } else {
            last.next = first;
            first.prev = last;
        }
    }
}

小心使用toString方法，即使是调试也非常重要！

public String toString(){
     String toReturn = "[ " + this.data;

     Node current=this.next;
     while (current!=null && current != this){ // <- cycle protection
         toReturn+=" -> " + current.data;
         current=current.next;
     }
     toReturn+=" ]";
     return toReturn;
 }

修改

它总是返回头节点：

public static Node reverse(Node head, Node tail) { Node first = head; Node last = tail; while (first != last && first.prev != last) { Node preFirst = first.prev; Node postFirst = first.next; Node preLast = last.prev; Node postLast = last.next; if (preFirst != null) { preFirst.next = last; } if (postLast != null) { postLast.prev = first; } last.prev = preFirst; first.next = postLast; if (last != postFirst){ last.next = postFirst; postFirst.prev = last; first.prev = preLast; preLast.next = first; first = postFirst; last = preLast; } else { last.next = first; first.prev = last; } } Node result = tail; while (result.prev != null){ result = result.prev; } return result; }

示例：

public static void main(String... args) { Node one = new Node(1); Node two = new Node(2); Node three = new Node(3); Node four = new Node(4); Node five = new Node(5); five.setNodes(four, null); four.setNodes(three, five); three.setNodes(two, four); two.setNodes(one, three); one.setNodes(null, two); System.out.println("before reversing..."); System.out.println(one); System.out.println("After reversing..."); one = reverse(one, three); System.out.println(one); }

打印：

before reversing... [ 1 -> 2 -> 3 -> 4 -> 5 ] After reversing... [ 3 -> 2 -> 1 -> 4 -> 5 ]

双链表中给定节点之间的反向链表 - 算法

1 个答案: