用于打印链表的反转而不实际反转的 Java 程序

原文:https://www . geesforgeks . org/Java-打印程序-链表的反向而不实际反向/

给定一个链表，使用递归函数打印它的逆序。例如，如果给定的链表是 1->2->3->4，那么输出应该是 4->3->2->1。 注意，问题只是关于打印反面。要反转列表本身请参见本T3难度等级:菜鸟

reverse-a-link-list

算法:

printReverse(head)
  1\. call print reverse for head->next
  2\. print head->data

实施:

Java 语言(一种计算机语言，尤用于创建网站)

// Java program to print reverse 
// of a linked list
class LinkedList
{
    // Head of list
    Node head;  

    // Linked list Node
    class Node
    {
        int data;
        Node next;
        Node(int d) 
        {
            data = d; 
            next = null; 
        }
    }

    // Function to print reverse of 
    // linked list 
    void printReverse(Node head)
    {
        if (head == null) return;

        // Print list of head node
        printReverse(head.next);

        // After everything else is printed,
        //  Print head
        System.out.print(head.data + " ");
    }

    // Utility Functions 

    // Inserts a new Node at front 
    // of the list. 
    public void push(int new_data)
    {
        /* 1 & 2: Allocate the Node &
                  Put in the data*/
        Node new_node = new Node(new_data);

        // 3\. Make next of new Node as head 
        new_node.next = head;

        // 4\. Move the head to point 
        // to new Node 
        head = new_node;
    }

    // Driver code
    public static void main(String args[])
    {
        // Create linked list 1->2->3->4
        LinkedList llist = new LinkedList();
        llist.push(4);
        llist.push(3);
        llist.push(2);
        llist.push(1);

        llist.printReverse(llist.head);
    }
}
// This code is contributed by Rajat Mishra 

输出:

4 3 2 1

时间复杂度: O(n)

更多详情请参考打印链表的反转而不实际反转的完整文章！

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