算法设计:修改链表的内容

2021年5月3日17:09:08 发表评论 780 次浏览

本文概述

给定一个包含以下内容的单链表

n个节点。修改前半节点的值, 以使第一个节点的新值等于最后一个节点的值减去第一个节点的当前值, 第二个节点的新值等于后一个节点的值减去第二个节点的当前值, 对于前半节点也是如此。如果n如果为奇数, 则中间节点的值保持不变。

(没有多余的内存可以使用)。

例子:

Input : 10 -> 4 -> 5 -> 3 -> 6
Output : 4 -> 1 -> 5 -> 3 -> 6

Input : 2 -> 9 -> 8 -> 12 -> 7 -> 10
Output : -8 -> 2 -> -4 -> 12 -> 7 -> 10

在亚马逊采访中问

方法:以下步骤是:

  1. 从中间拆分列表。执行前后拆分。如果元素的数量为奇数, 则多余的元素应放在第一(前)列表中。
  2. 反转第二个(后退)列表.
  3. 同时遍历两个列表时, 请执行所需的减法操作。
  4. 再次反转第二个列表。
  5. 将第二个列表连接回第一个列表的末尾。

C ++

//C++ implementation to modify the contents of 
//the linked list
#include <bits/stdc++.h>
using namespace std;
  
/* Linked list node */
struct Node
{
     int data;
     struct Node* next;
};
  
/* function prototype for printing the list */
void printList( struct Node*);
  
/* Function to insert a node at the beginning of 
    the linked 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 = new_data;
    
   /* link the old list at the end of the new node */
   new_node->next = *head_ref;    
    
   /* move the head to point to the new node */
   *head_ref = new_node;
} 
  
/* Split the nodes of the given list 
    into front and back halves, and return the two lists 
    using the reference parameters.
    Uses the fast/slow pointer strategy. */
void frontAndBackSplit( struct Node *head, struct Node **front_ref, struct Node **back_ref)
{
     Node *slow, *fast;
      
     slow = head;
     fast = head->next;
      
     /* Advance 'fast' two nodes, and 
        advance 'slow' one node */
     while (fast != NULL)
     {
         fast = fast->next;
         if (fast != NULL)
         {
             slow = slow->next;
             fast = fast->next;
         }
     }
      
      /* 'slow' is before the midpoint in the list, so split it in two at that point. */
     *front_ref = head;
     *back_ref = slow->next;
     slow->next = NULL;
}
  
/* Function to reverse the linked list */
void reverseList( struct Node **head_ref)
{
     struct Node *current, *prev, *next;
     current = *head_ref;
     prev = NULL;
     while (current != NULL)
     {
         next = current->next;
         current->next = prev;
         prev = current;
         current = next;
     }    
     *head_ref = prev;
}
  
//perfrom the required subtraction operation on
//the 1st half of the linked list
void modifyTheContentsOf1stHalf( struct Node *front, struct Node *back)
{
     //traversing both the lists simultaneously
     while (back != NULL)
     {
         //subtraction operation and node data
         //modification
         front->data = front->data - back->data;
          
         front = front->next;
         back = back->next;
     }
}
  
//function to concatenate the 2nd(back) list at the end of
//the 1st(front) list and returns the head of the new list
struct Node* concatFrontAndBackList( struct Node *front, struct Node *back)
{
     struct Node *head = front;
      
     while (front->next != NULL)
         front = front->next;    
          
     front->next    = back;
      
     return head;
}
  
//function to modify the contents of the linked list
struct Node* modifyTheList( struct Node *head)
{
     //if list is empty or contains only single node
     if (!head || head->next == NULL)
         return head;
      
     struct Node *front, *back;
      
     //split the list into two halves
     //front and back lists
     frontAndBackSplit(head, &front, &back);    
          
     //reverse the 2nd(back) list
     reverseList(&back);
      
     //modify the contents of 1st half    
     modifyTheContentsOf1stHalf(front, back);
          
     //agains reverse the 2nd(back) list
     reverseList(&back);
      
     //concatenating the 2nd list back to the 
     //end of the 1st list
     head = concatFrontAndBackList(front, back);
      
     //pointer to the modified list
     return head;
}
  
//function to print the linked list
void printList( struct Node *head)
{
     if (!head)
         return ;
      
     while (head->next != NULL)
     {
         cout <<head->data <<" -> " ;
         head = head->next;
     }
     cout <<head->data <<endl;
}
  
//Driver program to test above
int main()
{
     struct Node *head = NULL;
      
     //creating the linked list
     push(&head, 10);
     push(&head, 7);
     push(&head, 12);
     push(&head, 8);
     push(&head, 9);
     push(&head, 2);
      
     //modify the linked list
     head = modifyTheList(head);
      
     //print the modified linked list
     cout <<"Modified List:" <<endl;
     printList(head);
     return 0;
}

Java

//Java implementation to modify the contents 
//of the linked list
class GFG
{
      
/* Linked list node */
static class Node
{
     int data;
     Node next;
};
  
/* Function to insert a node at the beginning 
of the linked list */
static Node push(Node head_ref, int new_data)
{
     /* allocate node */
     Node new_node = new Node();
     /* put in the data */
     new_node.data = new_data;
          
     /* link the old list at the end 
     of the new node */
     new_node.next = head_ref; 
          
     /* move the head to point to the new node */
     head_ref = new_node;
      
     return head_ref;
} 
  
static Node front, back;
  
/* Split the nodes of the given list 
into front and back halves, and return the two lists 
using the reference parameters.
Uses the fast/slow pointer strategy. */
static void frontAndBackSplit( Node head)
{
     Node slow, fast;
      
     slow = head;
     fast = head.next;
      
     /* Advance 'fast' two nodes, and 
     advance 'slow' one node */
     while (fast != null )
     {
         fast = fast.next;
         if (fast != null )
         {
             slow = slow.next;
             fast = fast.next;
         }
     }
      
     /* 'slow' is before the midpoint in the list, so split it in two at that point. */
     front = head;
     back = slow.next;
     slow.next = null ;
}
  
/* Function to reverse the linked list */
static Node reverseList( Node head_ref)
{
     Node current, prev, next;
     current = head_ref;
     prev = null ;
     while (current != null )
     {
         next = current.next;
         current.next = prev;
         prev = current;
         current = next;
     } 
     head_ref = prev;
     return head_ref;
}
  
//perfrom the required subtraction operation 
//on the 1st half of the linked list
static void modifyTheContentsOf1stHalf()
{
     Node front1 = front, back1 = back;
     //traversing both the lists simultaneously
     while (back1 != null )
     {
         //subtraction operation and node data
         //modification
         front1.data = front1.data - back1.data;
          
         front1 = front1.next;
         back1 = back1.next;
     }
}
  
//function to concatenate the 2nd(back) list 
//at the end of the 1st(front) list and 
//returns the head of the new list
static Node concatFrontAndBackList(Node front, Node back)
{
     Node head = front;
      
     if (front == null ) return back;
      
     while (front.next != null )
         front = front.next; 
          
     front.next = back;
      
     return head;
}
  
//function to modify the contents of the linked list
static Node modifyTheList( Node head)
{
     //if list is empty or contains only single node
     if (head == null || head.next == null )
         return head;
     front = null ; back = null ;
      
     //split the list into two halves
     //front and back lists
     frontAndBackSplit(head);
          
     //reverse the 2nd(back) list
     back = reverseList(back);
      
     //modify the contents of 1st half 
     modifyTheContentsOf1stHalf();
      
     //agains reverse the 2nd(back) list
     back = reverseList(back);
      
     //concatenating the 2nd list back to the 
     //end of the 1st list
     head = concatFrontAndBackList(front, back);
  
     //pointer to the modified list
     return head;
}
  
//function to print the linked list
static void printList( Node head)
{
     if (head == null )
         return ;
      
     while (head.next != null )
     {
         System.out.print(head.data + " -> " );
         head = head.next;
     }
     System.out.println(head.data );
}
  
//Driver Code
public static void main(String args[])
{
     Node head = null ;
      
     //creating the linked list
     head = push(head, 10 );
     head = push(head, 7 );
     head = push(head, 12 );
     head = push(head, 8 );
     head = push(head, 9 );
     head = push(head, 2 );
      
     //modify the linked list
     head = modifyTheList(head);
      
     //print the modified linked list
     System.out.println( "Modified List:" );
     printList(head);
}
}
  
//This code is contributed by Arnab Kundu

python

# Python implementation to modify the contents 
# of the linked list
  
# Linked list node 
class Node: 
      
     def __init__( self , data): 
         self .data = data 
         self . next = None
  
# Function to insert a node at the beginning 
# of the linked list 
def push(head_ref, new_data):
  
     # allocate node 
     new_node = Node( 0 )
      
     # put in the data 
     new_node.data = new_data
          
     # link the old list at the end 
     #of the new node 
     new_node. next = head_ref 
          
     # move the head to point to the new node 
     head_ref = new_node
      
     return head_ref
  
front = None
back = None
  
# Split the nodes of the given list 
# into front and back halves, # and return the two lists 
# using the reference parameters.
# Uses the fast/slow pointer strategy. 
def frontAndBackSplit( head):
  
     global front
     global back
     slow = None
     fast = None
      
     slow = head
     fast = head. next
      
     # Advance 'fast' two nodes, and 
     # advance 'slow' one node 
     while (fast ! = None ):
      
         fast = fast. next
         if (fast ! = None ):
             slow = slow. next
             fast = fast. next
  
     # 'slow' is before the midpoint in the list, # so split it in two at that point. 
     front = head
     back = slow. next
     slow. next = None
     return head
  
# Function to reverse the linked list 
def reverseList( head_ref):
  
     current = None
     prev = None
     next = None
     current = head_ref
     prev = None
     while (current ! = None ):
      
         next = current. next
         current. next = prev
         prev = current
         current = next
      
     head_ref = prev
     return head_ref
  
# perfrom the required subtraction operation 
# on the 1st half of the linked list
def modifyTheContentsOf1stHalf():
  
     global front
     global back
     front1 = front
     back1 = back
      
     # traversing both the lists simultaneously
     while (back1 ! = None ):
      
         # subtraction operation and node data
         # modification
         front1.data = front1.data - back1.data
          
         front1 = front1. next
         back1 = back1. next
      
# function to concatenate the 2nd(back) list 
# at the end of the 1st(front) list and 
# returns the head of the new list
def concatFrontAndBackList( front, back):
      
     head = front
      
     if (front = = None ):
         return back
      
     while (front. next ! = None ):
         front = front. next
          
     front. next = back
     return head
  
# function to modify the contents of the linked list
def modifyTheList( head):
  
     global front
     global back
  
     # if list is empty or contains only single node
     if (head = = None or head. next = = None ):
         return head
     front = None
     back = None
      
     # split the list into two halves
     # front and back lists
     frontAndBackSplit(head)
          
     # reverse the 2nd(back) list
     back = reverseList(back)
      
     # modify the contents of 1st half 
     modifyTheContentsOf1stHalf()
      
     # agains reverse the 2nd(back) list
     back = reverseList(back)
      
     # concatenating the 2nd list back to the 
     # end of the 1st list
     head = concatFrontAndBackList(front, back)
  
     # pointer to the modified list
     return head
  
# function to print the linked list
def printList( head):
  
     if (head = = None ):
         return
      
     while (head. next ! = None ):
      
         print (head.data , " -> " , end = "")
         head = head. next
      
     print (head.data )
  
# Driver Code
  
head = None
      
# creating the linked list
head = push(head, 10 )
head = push(head, 7 )
head = push(head, 12 )
head = push(head, 8 )
head = push(head, 9 )
head = push(head, 2 )
      
# modify the linked list
head = modifyTheList(head)
      
# print the modified linked list
print ( "Modified List:" )
printList(head)
  
# This code is contributed by Arnab Kundu

C#

//C# implementation to modify the 
//contents of the linked list
using System;
  
class GFG
{
      
/* Linked list node */
public class Node
{
     public int data;
     public Node next;
};
  
/* Function to insert a node at 
the beginning of the linked list */
static Node push(Node head_ref, int new_data)
{
     /* allocate node */
     Node new_node = new Node();
      
     /* put in the data */
     new_node.data = new_data;
          
     /* link the old list at the end 
     of the new node */
     new_node.next = head_ref; 
          
     /* move the head to point to the new node */
     head_ref = new_node;
      
     return head_ref;
} 
  
static Node front, back;
  
/* Split the nodes of the given list 
into front and back halves, and return the two lists 
using the reference parameters.
Uses the fast/slow pointer strategy. */
static void frontAndBackSplit( Node head)
{
     Node slow, fast;
      
     slow = head;
     fast = head.next;
      
     /* Advance 'fast' two nodes, and 
     advance 'slow' one node */
     while (fast != null )
     {
         fast = fast.next;
         if (fast != null )
         {
             slow = slow.next;
             fast = fast.next;
         }
     }
      
     /* 'slow' is before the midpoint in the list, so split it in two at that point. */
     front = head;
     back = slow.next;
     slow.next = null ;
}
  
/* Function to reverse the linked list */
static Node reverseList(Node head_ref)
{
     Node current, prev, next;
     current = head_ref;
     prev = null ;
     while (current != null )
     {
         next = current.next;
         current.next = prev;
         prev = current;
         current = next;
     } 
     head_ref = prev;
     return head_ref;
}
  
//perfrom the required subtraction operation 
//on the 1st half of the linked list
static void modifyTheContentsOf1stHalf()
{
     Node front1 = front, back1 = back;
      
     //traversing both the lists simultaneously
     while (back1 != null )
     {
         //subtraction operation and node data
         //modification
         front1.data = front1.data - back1.data;
          
         front1 = front1.next;
         back1 = back1.next;
     }
}
  
//function to concatenate the 2nd(back) list 
//at the end of the 1st(front) list and 
//returns the head of the new list
static Node concatFrontAndBackList(Node front, Node back)
{
     Node head = front;
      
     if (front == null )
         return back;
      
     while (front.next != null )
         front = front.next; 
          
     front.next = back;
      
     return head;
}
  
//function to modify the contents of
//the linked list
static Node modifyTheList(Node head)
{
     //if list is empty or contains 
     //only single node
     if (head == null || head.next == null )
         return head;
     front = null ; back = null ;
      
     //split the list into two halves
     //front and back lists
     frontAndBackSplit(head);
          
     //reverse the 2nd(back) list
     back = reverseList(back);
      
     //modify the contents of 1st half 
     modifyTheContentsOf1stHalf();
      
     //agains reverse the 2nd(back) list
     back = reverseList(back);
      
     //concatenating the 2nd list back to the 
     //end of the 1st list
     head = concatFrontAndBackList(front, back);
  
     //pointer to the modified list
     return head;
}
  
//function to print the linked list
static void printList( Node head)
{
     if (head == null )
         return ;
      
     while (head.next != null )
     {
         Console.Write(head.data + " -> " );
         head = head.next;
     }
     Console.WriteLine(head.data );
}
  
//Driver Code
public static void Main()
{
     Node head = null ;
      
     //creating the linked list
     head = push(head, 10);
     head = push(head, 7);
     head = push(head, 12);
     head = push(head, 8);
     head = push(head, 9);
     head = push(head, 2);
      
     //modify the linked list
     head = modifyTheList(head);
      
     //print the modified linked list
     Console.WriteLine( "Modified List:" );
     printList(head);
}
}
  
//This code is contributed by PrinciRaj1992

输出如下:

Modified List:
-8 -> 2 -> -4 -> 12 -> 7 -> 10

时间复杂度:O(n), 其中ñ在节点数上。

另一种方法(使用堆栈):

1.找到下半个链表的起点。

2.将后半列表的所有元素推入堆栈s中。

3.使用temp从头开始遍历列表, 直到堆栈不为空

并通过减去每个节点的堆栈顶部元素来修改temp-> data。

下面是使用堆栈的实现。

C ++

//C++ implementation to modify the
//contents of the linked list
#include <bits/stdc++.h>
using namespace std;
  
//Linked list node
struct Node
{
     int data;
     struct Node* next;
};
  
//function prototype for printing the list
void printList( struct Node*);
  
//Function to insert a node at the
//beginning of the linked 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 = new_data;
  
//link the old list at the end of the new node
new_node->next = *head_ref; 
  
//move the head to point to the new node
*head_ref = new_node;
} 
  
//function to print the linked list
void printList( struct Node *head)
{
     if (!head)
         return ;
      
     while (head->next != NULL)
     {
         cout <<head->data <<" -> " ;
         head = head->next;
     }
     cout <<head->data <<endl;
}
  
//Function to middle node of list.
Node* find_mid(Node *head)
{
     Node *temp = head, *slow = head, *fast = head ;
      
     while (fast && fast->next)
     {
          
     //Advance 'fast' two nodes, and 
     //advance 'slow' one node
     slow = slow->next ;
     fast = fast->next->next ;
     }
      
     //If number of nodes are odd then update slow
     //by slow->next;
     if (fast)
     slow = slow->next ;
  
return slow ;
}
  
//function to modify the contents of the linked list.
void modifyTheList( struct Node *head, struct Node *slow)
{
//Create Stack. 
stack <int> s;
Node *temp = head ;
  
while (slow)
{
     s.push( slow->data ) ;
     slow = slow->next ;
}
  
//Traverse the list by using temp until stack is empty.
while ( !s.empty() )
{
     temp->data = temp->data - s.top() ;
     temp = temp->next ;
     s.pop() ;
}
  
}
  
//Driver program to test above
int main()
{
     struct Node *head = NULL, *mid ;
      
     //creating the linked list
     push(&head, 10);
     push(&head, 7);
     push(&head, 12);
     push(&head, 8);
     push(&head, 9);
     push(&head, 2);
      
     //Call Function to Find the starting point of second half of list. 
     mid = find_mid(head) ;
      
     //Call function to modify the contents of the linked list.
     modifyTheList( head, mid);
      
      
     //print the modified linked list
     cout <<"Modified List:" <<endl;
     printList(head);
     return 0;
} 
  
//This is contributed by Mr. Gera

Java

//Java implementation to modify the
//contents of the linked list
import java.util.*;
  
class GFG 
{
  
     //Linked list node
     static class Node 
     {
  
         int data;
         Node next;
     };
  
     //Function to insert a node at the
     //beginning of the linked list
     static Node push(Node head_ref, int new_data) 
     {
  
         //allocate node
         Node new_node = new Node();
  
         //put in the data
         new_node.data = new_data;
  
         //link the old list at the end of the new node
         new_node.next = head_ref;
  
         //move the head to point to the new node
         head_ref = new_node;
         return head_ref;
     }
  
     //function to print the linked list
     static void printList(Node head) 
     {
         if (head == null ) 
         {
             return ;
         }
  
         while (head.next != null ) 
         {
             System.out.print(head.data + "->" );
             head = head.next;
         }
         System.out.print(head.data + "\n" );
     }
  
     //Function to middle node of list.
     static Node find_mid(Node head) 
     {
         Node temp = head, slow = head, fast = head;
  
         while (fast != null && fast.next != null )
         {
  
             //Advance 'fast' two nodes, and 
             //advance 'slow' one node
             slow = slow.next;
             fast = fast.next.next;
         }
  
         //If number of nodes are odd then update slow
         //by slow.next;
         if (fast != null ) 
         {
             slow = slow.next;
         }
  
         return slow;
     }
  
     //function to modify the contents of the linked list.
     static void modifyTheList(Node head, Node slow) 
     {
         //Create Stack. 
         Stack<Integer> s = new Stack<Integer>();
         Node temp = head;
  
         while (slow != null ) 
         {
             s.add(slow.data);
             slow = slow.next;
         }
  
     //Traverse the list by using temp until stack is empty.
         while (!s.empty())
         {
             temp.data = temp.data - s.peek();
             temp = temp.next;
             s.pop();
         }
  
     }
  
     //Driver program to test above
     public static void main(String[] args)
     {
         Node head = null , mid;
  
         //creating the linked list
         head = push(head, 10 );
         head = push(head, 7 );
         head = push(head, 12 );
         head = push(head, 8 );
         head = push(head, 9 );
         head = push(head, 2 );
  
         //Call Function to Find the starting
         //point of second half of list. 
         mid = find_mid(head);
  
         //Call function to modify 
         //the contents of the linked list.
         modifyTheList(head, mid);
  
         //print the modified linked list
         System.out.print( "Modified List:" + "\n" );
         printList(head);
     }
}
  
//This code is contributed by Rajput-Ji

C#

//C# implementation to modify the
//contents of the linked list
using System;
using System.Collections.Generic;
  
class GFG 
{
  
     //Linked list node
     public class Node 
     {
  
         public int data;
         public Node next;
     };
  
     //Function to insert a node at the
     //beginning of the linked list
     static Node push(Node head_ref, int new_data) 
     {
  
         //allocate node
         Node new_node = new Node();
  
         //put in the data
         new_node.data = new_data;
  
         //link the old list at the end of the new node
         new_node.next = head_ref;
  
         //move the head to point to the new node
         head_ref = new_node;
         return head_ref;
     }
  
     //function to print the linked list
     static void printList(Node head) 
     {
         if (head == null ) 
         {
             return ;
         }
  
         while (head.next != null ) 
         {
             Console.Write(head.data + "->" );
             head = head.next;
         }
         Console.Write(head.data + "\n" );
     }
  
     //Function to middle node of list.
     static Node find_mid(Node head) 
     {
         Node temp = head, slow = head, fast = head;
  
         while (fast != null && fast.next != null )
         {
  
             //Advance 'fast' two nodes, and 
             //advance 'slow' one node
             slow = slow.next;
             fast = fast.next.next;
         }
  
         //If number of nodes are odd then update slow
         //by slow.next;
         if (fast != null ) 
         {
             slow = slow.next;
         }
  
         return slow;
     }
  
     //function to modify the contents of the linked list.
     static void modifyTheList(Node head, Node slow) 
     {
         //Create Stack. 
         Stack<int> s = new Stack<int>();
         Node temp = head;
  
         while (slow != null ) 
         {
             s.Push(slow.data);
             slow = slow.next;
         }
  
         //Traverse the list by using temp until stack is empty.
         while (s.Count != 0)
         {
             temp.data = temp.data - s.Peek();
             temp = temp.next;
             s.Pop();
         }
  
     }
  
     //Driver code
     public static void Main(String[] args)
     {
         Node head = null , mid;
  
         //creating the linked list
         head = push(head, 10);
         head = push(head, 7);
         head = push(head, 12);
         head = push(head, 8);
         head = push(head, 9);
         head = push(head, 2);
  
         //Call Function to Find the starting
         //point of second half of list. 
         mid = find_mid(head);
  
         //Call function to modify 
         //the contents of the linked list.
         modifyTheList(head, mid);
  
         //print the modified linked list
         Console.Write( "Modified List:" + "\n" );
         printList(head);
     }
}
  
//This code is contributed by PrinciRaj1992

输出如下:

Modified List:
-8 -> 2 -> -4 -> 12 -> 7 -> 10

时间复杂度:O(n)

空间复杂度:O(n/2)

参考文献: https://www.careercup.com/question?id=5657550909341696

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