算法题（如何删除链表中的备用节点（详细实现）） _数据结构

本文概述

建议：在继续解决方案之前, 请先在"实践"上解决它。
C ++
C
Java
Python3
C#
C ++
C
Java
Python3
C#

给定一个单链列表, 从第二个节点开始删除它的所有备用节点。例如, 如果给定的链表为1-> 2-> 3-> 4-> 5, 则你的函数应将其转换为1-> 3-> 5, 并且给定的链表为1-> 2-> 3-> 4, 然后将其转换为1-> 3。
推荐：请在"实践首先, 在继续解决方案之前。【算法题（如何删除链表中的备用节点（详细实现））】方法1(迭代)
跟踪要删除的节点的上一个节点。首先更改上一个节点的下一个链接, 然后释放为该节点分配的内存。
C ++

// C++ program to remove alternate // nodes of a linked list #include < bits/stdc++.h> using namespace std; /* A linked list node */ class Node { public : int data; Node *next; }; /* deletes alternate nodes of a list starting with head */ void deleteAlt(Node *head) { if (head == NULL) return ; /* Initialize prev and node to be deleted */ Node *prev = head; Node *node = head-> next; while (prev != NULL & & node != NULL) { /* Change next link of previous node */ prev-> next = node-> next; /* Free memory */ free (node); /* Update prev and node */ prev = prev-> next; if (prev != NULL) node = prev-> next; } } /* UTILITY FUNCTIONS TO TEST fun1() and fun2() */ /* Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front of the list. */ void push(Node** head_ref, int new_data) { /* allocate node */ Node* new_node = new Node(); /* put in the data */ new_node-> data = https://www.lsbin.com/new_data; /* link the old list off the new node */ new_node-> next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node; } /* Function to print nodes in a given linked list */ void printList(Node *node) { while (node != NULL) { cout< < node-> data< <" " ; node = node-> next; } } /* Driver code */ int main() { /* Start with the empty list */ Node* head = NULL; /* Using push() to construct below list 1-> 2-> 3-> 4-> 5 */ push(& head, 5); push(& head, 4); push(& head, 3); push(& head, 2); push(& head, 1); cout< < "List before calling deleteAlt() \n" ; printList(head); deleteAlt(head); cout< < "\nList after calling deleteAlt() \n" ; printList(head); return 0; } // This code is contributed by rathbhupendra

// C program to remove alternate nodes of a linked list #include< stdio.h> #include< stdlib.h> /* A linked list node */ struct Node { int data; struct Node *next; }; /* deletes alternate nodes of a list starting with head */ void deleteAlt( struct Node *head) { if (head == NULL) return ; /* Initialize prev and node to be deleted */ struct Node *prev = head; struct Node *node = head-> next; while (prev != NULL & & node != NULL) { /* Change next link of previous node */ prev-> next = node-> next; /* Free memory */ free (node); /* Update prev and node */ prev = prev-> next; if (prev != NULL) node = prev-> next; } }/* UTILITY FUNCTIONS TO TEST fun1() and fun2() */ /* Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front of the list. */ void push( struct Node** head_ref, int new_data) { /* allocate node */ struct Node* new_node = ( struct Node*) malloc ( sizeof ( struct Node)); /* put in the data*/ new_node-> data= https://www.lsbin.com/new_data; /* link the old list off the new node */ new_node-> next = (*head_ref); /* move the head to point to the new node */ (*head_ref)= new_node; }/* Function to print nodes in a given linked list */ void printList( struct Node *node) { while (node != NULL) { printf ("%d " , node-> data); node = node-> next; } }/* Driver program to test above functions */ int main() { /* Start with the empty list */ struct Node* head = NULL; /* Using push() to construct below list 1-> 2-> 3-> 4-> 5*/ push(& head, 5); push(& head, 4); push(& head, 3); push(& head, 2); push(& head, 1); printf ( "\nList before calling deleteAlt() \n" ); printList(head); deleteAlt(head); printf ( "\nList after calling deleteAlt() \n" ); printList(head); return 0; }

Java

// Java program to delete alternate nodes of a linked list class LinkedList { Node head; // head of list/* Linked list Node*/ class Node { int data; Node next; Node( int d) {data = https://www.lsbin.com/d; next = null ; } }void deleteAlt() { if (head == null ) return ; Node prev = head; Node now = head.next; while (prev != null & & now != null ) { /* Change next link of previus node */ prev.next = now.next; /* Free node */ now = null ; /*Update prev and now */ prev = prev.next; if (prev != null ) now = prev.next; } }/* 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; }/* 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 program to test above functions */ public static void main(String args[]) { LinkedList llist = new LinkedList(); /* Constructed Linked List is 1-> 2-> 3-> 4-> 5-> null */ llist.push( 5 ); llist.push( 4 ); llist.push( 3 ); llist.push( 2 ); llist.push( 1 ); System.out.println( "Linked List before calling deleteAlt() " ); llist.printList(); llist.deleteAlt(); System.out.println( "Linked List after calling deleteAlt() " ); llist.printList(); } } /* This code is contributed by Rajat Mishra */

Python3

# Python3 program to remove alternate # nodes of a linked list import math # A linked list node class Node: def __init__( self , data): self .data = https://www.lsbin.com/data self . next = None# deletes alternate nodes # of a list starting with head def deleteAlt(head): if (head = = None ): return# Initialize prev and node to be deleted prev = head now = head. nextwhile (prev ! = None and now ! = None ): # Change next link of previous node prev. next = now. next# Free memory now = None# Update prev and node prev = prev. next if (prev ! = None ): now = prev. next# UTILITY FUNCTIONS TO TEST fun1() and fun2() # Given a reference (poer to poer) to the head # of a list and an , push a new node on the front # of the list. def push(head_ref, new_data): # allocate node new_node = Node(new_data)# put in the data new_node.data = new_data # link the old list off the new node new_node. next = head_ref # move the head to po to the new node head_ref = new_node return head_ref# Function to print nodes in a given linked list def prList(node): while (node ! = None ): print (node.data, end =" " ) node = node. next# Driver code if __name__ = = '__main__' : # Start with the empty list head = None# Using head=push() to construct below list # 1.2.3.4.5 head = push(head, 5 ) head = push(head, 4 ) head = push(head, 3 ) head = push(head, 2 ) head = push(head, 1 ) print ( "List before calling deleteAlt() " ) prList(head) deleteAlt(head) print ( "\nList after calling deleteAlt() " ) prList(head) # This code is contributed by Srathore

// C# program to delete alternate // nodes of a linked list using System; public class LinkedList { Node head; // head of list/* Linked list Node*/ public class Node { public int data; public Node next; public Node( int d) { data = https://www.lsbin.com/d; next = null ; } }void deleteAlt() { if (head == null ) return ; Node prev = head; Node now = head.next; while (prev != null & & now != null ) { /* Change next link of previus node */ prev.next = now.next; /* Free node */ now = null ; /*Update prev and now */ prev = prev.next; if (prev != null ) now = prev.next; } }/* 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; }/* Function to print linked list */ void printList() { Node temp = head; while (temp != null ) { Console.Write(temp.data+" " ); temp = temp.next; } Console.WriteLine(); }/* Driver code*/ public static void Main(String []args) { LinkedList llist = new LinkedList(); /* Constructed Linked List is 1-> 2-> 3-> 4-> 5-> null */ llist.push(5); llist.push(4); llist.push(3); llist.push(2); llist.push(1); Console.WriteLine( "Linked List before" + "calling deleteAlt() " ); llist.printList(); llist.deleteAlt(); Console.WriteLine( "Linked List after" + "calling deleteAlt() " ); llist.printList(); } }// This code has been contributed // by 29AjayKumar

输出如下：

List before calling deleteAlt() 1 2 3 4 5 List after calling deleteAlt() 1 3 5

时间复杂度：O(n)其中n是给定链表中节点的数量。
方法2(递归)
递归代码使用与方法1相同的方法。递归代码很简单且简短, 但是导致O(n)递归函数调用大小为n的链表。
C ++

/* deletes alternate nodes of a list starting with head */ void deleteAlt(Node *head) { if (head == NULL) return ; Node *node = head-> next; if (node == NULL) return ; /* Change the next link of head */ head-> next = node-> next; /* free memory allocated for node */ free (node); /* Recursively call for the new next of head */ deleteAlt(head-> next); } // This code is contributed by rathbhupendra

/* deletes alternate nodes of a list starting with head */ void deleteAlt( struct Node *head) { if (head == NULL) return ; struct Node *node = head-> next; if (node == NULL) return ; /* Change the next link of head */ head-> next = node-> next; /* free memory allocated for node */ free (node); /* Recursively call for the new next of head */ deleteAlt(head-> next); }

Java

/* deletes alternate nodes of a list starting with head */ static Node deleteAlt(Node head) { if (head == null ) return ; Node node = head.next; if (node == null ) return ; /* Change the next link of head */ head.next = node.next; /* Recursively call for the new next of head */ head.next = deleteAlt(head.next); } // This code is contributed by Arnab Kundu

Python3

# deletes alternate nodes of a list starting with head def deleteAlt(head): if (head = = None ): returnnode = head. nextif (node = = None ): return# Change the next link of head head. next = node. next# free memory allocated for node #free(node) # Recursively call for the new next of head deleteAlt(head. next ) # This code is contributed by Srathore

/* deletes alternate nodes of a list starting with head */ static Node deleteAlt(Node head) { if (head == null ) return ; Node node = head.next; if (node == null ) return ; /* Change the next link of head */ head.next = node.next; /* Recursively call for the new next of head */ head.next = deleteAlt(head.next); } // This code is contributed by Arnab Kundu

时间复杂度：O(n)
如果你发现上述代码/算法有误, 请写注释, 或者找到解决同一问题的更好方法。