删除右侧值较大的节点的 Java 程序
原文:https://www . geesforgeks . org/Java-program-to-delete-nodes-哪些节点右侧的值更大/
给定一个单链表，移除右边值较大的所有节点。
示例:
Input: 12->15->10->11->5->6->2->3->NULL
Output: 15->11->6->3->NULL
Explanation: 12, 10, 5 and 2 have been deleted because there is a
greater value on the right side. When we examine 12,
we see that after 12 there is one node with a value
greater than 12 (i.e. 15), so we delete 12. When we
examine 15, we find no node after 15 that has a value
greater than 15, so we keep this node. When we go like
this, we get 15->6->3
Input: 10->20->30->40->50->60->NULL
Output: 60->NULL
Explanation: 10, 20, 30, 40, and 50 have been deleted because
they all have a greater value on the right side.
Input: 60->50->40->30->20->10->NULL
Output: No Change.
方法 1(简单): 使用两个循环。在外循环中，逐个挑选链表的节点。在内部循环中，检查是否存在其值大于拾取节点的节点。如果存在值更大的节点，则删除拾取的节点。 时间复杂度: O(n^2)
方法 2(反向使用): 感谢帕拉斯提供以下算法。 1。颠倒列表。 2。遍历反向列表。把麦克斯留到现在。如果下一个节点小于 max，则删除下一个节点，否则 max =下一个节点。 3。再次反转列表以保留原始顺序。 时间复杂度: O(n) 感谢 R.Srinivasan 提供下面的代码。
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to delete nodes which have
// a greater value on right side
class LinkedList
{
// head of list
Node head;
// Linked list Node
class Node
{
int data;
Node next;
Node(int d)
{
data = d;
next = null;
}
}
/* Deletes nodes which have a node
with greater value node on left side */
void delLesserNodes()
{
// 1.Reverse the linked list
reverseList();
/* 2\. In the reversed list, delete nodes
which have a node with greater value
node on left side. Note that head node
is never deleted because it is the
leftmost node.*/
_delLesserNodes();
/* 3\. Reverse the linked list again to retain
the original order */
reverseList();
}
/* Deletes nodes which have greater value
node(s) on left side */
void _delLesserNodes()
{
Node current = head;
// Initialise max
Node maxnode = head;
Node temp;
while (current != null &&
current.next != null)
{
/* If current is smaller than max,
then delete current */
if (current.next.data < maxnode.data)
{
temp = current.next;
current.next = temp.next;
temp = null;
}
/* If current is greater than max,
then update max and move current */
else
{
current = current.next;
maxnode = current;
}
}
}
// Utility functions
/* Inserts a new Node at front of
the list. */
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;
}
// Function to reverse the linked list
void reverseList()
{
Node current = head;
Node prev = null;
Node next;
while (current != null)
{
next = current.next;
current.next = prev;
prev = current;
current = next;
}
head = prev;
}
// Function to print linked list
void printList()
{
Node temp = head;
while (temp != null)
{
System.out.print(temp.data +
" ");
temp = temp.next;
}
System.out.println();
}
// Driver code
public static void main(String args[])
{
LinkedList llist = new LinkedList();
/* Constructed Linked List is
12->15->10->11->5->6->2->3 */
llist.push(3);
llist.push(2);
llist.push(6);
llist.push(5);
llist.push(11);
llist.push(10);
llist.push(15);
llist.push(12);
System.out.println("Given Linked List");
llist.printList();
llist.delLesserNodes();
System.out.println("Modified Linked List");
llist.printList();
}
}
// This code is contributed by Rajat Mishra
输出:
Given Linked List
12 15 10 11 5 6 2 3
Modified Linked List
15 11 6 3
方法 3: 另一个比较简单的方法是从开始遍历列表，当当前 node <下一个 Node 时删除该节点。要删除当前节点，请遵循这种方法。让我们假设您必须删除当前节点 X:
- 将下一个节点的数据复制到 X 中，即 X.data = X.next.data
- 复制下一个节点的下一个地址，即 X.next = X.next.next
仅当当前节点>下一个节点时，才在列表中向前移动。
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for above approach
import java.io.*;
// This class represents a single node
// in a linked list
class Node
{
int data;
Node next;
public Node(int data)
{
this.data = data;
this.next = null;
}
}
//This is a utility class for
// linked list
class LLUtil
{
// This function creates a linked list
// from a given array and returns head
public Node createLL(int[] arr)
{
Node head = new Node(arr[0]);
Node temp = head;
Node newNode = null;
for(int i = 1; i < arr.length; i++)
{
newNode = new Node(arr[i]);
temp.next = newNode;
temp = temp.next;
}
return head;
}
// This function prints given
// linked list
public void printLL(Node head)
{
while(head != null)
{
System.out.print(head.data +
" ");
head = head.next;
}
System.out.println();
}
}
// Driver code
class GFG
{
public static void main (String[] args)
{
int[] arr = {12, 15, 10, 11,
5, 6, 2, 3};
LLUtil llu = new LLUtil();
Node head = llu.createLL(arr);
System.out.println("Given Linked List");
llu.printLL(head);
head = deleteNodesOnRightSide(head);
System.out.println("Modified Linked List");
llu.printLL(head);
}
//Main function
public static Node deleteNodesOnRightSide(Node head)
{
if(head == null || head.next == null)
return head;
Node nextNode = deleteNodesOnRightSide(head.next);
if(nextNode.data > head.data)
return nextNode;
head.next = nextNode;
return head;
}
}
输出:
Given Linked List
12 15 10 11 5 6 2 3
Modified Linked List
15 11 6 3
详情请参考整篇文章删除右侧数值较大的节点！
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