使用任意指针指向链表中下一个更高值的节点
给定单链表,每个节点都有一个额外的“任意”指针,当前指向 NULL。需要使“任意”指针指向下一个更高值的节点。
我们强烈建议您将浏览器最小化并先尝试一下
一个简单的解决方案是一个一个地遍历所有节点,对于每个节点,找到当前节点下一个更大值的节点并改变下一个指针。此解决方案的时间复杂度为 O(n 2 )。
一个有效的解决方案在 O(nLogn) 时间内有效。这个想法是对链表使用合并排序。
1) 遍历输入列表并将下一个指针复制到每个节点的仲裁指针。
2) 对仲裁指针形成的链表进行归并排序。
下面是上述想法的实现。所有合并排序功能均取自此处。此处修改了所采用的函数,以便它们处理仲裁指针而不是下一个指针。
C++
// C++ program to populate arbit pointers
// to next higher value using merge sort
#include
using namespace std;
/* Link list node */
class Node
{
public:
int data;
Node* next, *arbit;
};
/* function prototypes */
Node* SortedMerge(Node* a, Node* b);
void FrontBackSplit(Node* source,
Node** frontRef, Node** backRef);
/* sorts the linked list formed by arbit pointers
(does not change next pointer or data) */
void MergeSort(Node** headRef)
{
Node* head = *headRef;
Node* a, *b;
/* Base case -- length 0 or 1 */
if ((head == NULL) || (head->arbit == NULL))
return;
/* Split head into 'a' and 'b' sublists */
FrontBackSplit(head, &a, &b);
/* Recursively sort the sublists */
MergeSort(&a);
MergeSort(&b);
/* answer = merge the two sorted lists together */
*headRef = SortedMerge(a, b);
}
/* See https://www.geeksforgeeks.org/?p=3622 for
details of this function */
Node* SortedMerge(Node* a, Node* b)
{
Node* result = NULL;
/* Base cases */
if (a == NULL)
return (b);
else if (b == NULL)
return (a);
/* Pick either a or b, and recur */
if (a->data <= b->data)
{
result = a;
result->arbit = SortedMerge(a->arbit, b);
}
else
{
result = b;
result->arbit = SortedMerge(a, b->arbit);
}
return (result);
}
/* Split the nodes of the given list into front
and back halves, and return the two lists using
the reference parameters. If the length is odd,
the extra node should go in the front list.
Uses the fast/slow pointer strategy. */
void FrontBackSplit(Node* source,
Node** frontRef, Node** backRef)
{
Node* fast, *slow;
if (source == NULL || source->arbit == NULL)
{
/* length < 2 cases */
*frontRef = source;
*backRef = NULL;
return;
}
slow = source, fast = source->arbit;
/* Advance 'fast' two nodes, and
advance 'slow' one node */
while (fast != NULL)
{
fast = fast->arbit;
if (fast != NULL)
{
slow = slow->arbit;
fast = fast->arbit;
}
}
/* 'slow' is before the midpoint in the list,
so split it in two at that point. */
*frontRef = source;
*backRef = slow->arbit;
slow->arbit = NULL;
}
/* Function to insert a node at the
beginning of the linked list */
void 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 off the new node */
new_node->next = (*head_ref);
new_node->arbit = NULL;
/* move the head to point to the new node */
(*head_ref) = new_node;
}
// Utility function to print result linked list
void printListafter(Node *node, Node *anode)
{
cout<<"Traversal using Next Pointer\n";
while (node!=NULL)
{
cout << node->data << ", ";
node = node->next;
}
printf("\nTraversal using Arbit Pointer\n");
while (anode!=NULL)
{
cout << anode->data << ", ";
anode = anode->arbit;
}
}
// This function populates arbit pointer in every node to the
// next higher value. And returns pointer to the node with
// minimum value
Node* populateArbit(Node *head)
{
// Copy next pointers to arbit pointers
Node *temp = head;
while (temp != NULL)
{
temp->arbit = temp->next;
temp = temp->next;
}
// Do merge sort for arbitrary pointers
MergeSort(&head);
// Return head of arbitrary pointer linked list
return head;
}
/* Driver program to test above functions*/
int main()
{
/* Start with the empty list */
Node* head = NULL;
/* Let us create the list shown above */
push(&head, 3);
push(&head, 2);
push(&head, 10);
push(&head, 5);
/* Sort the above created Linked List */
Node *ahead = populateArbit(head);
cout << "Result Linked List is: \n";
printListafter(head, ahead);
return 0;
}
// This is code is contributed by rathbhupendra
C
// C program to populate arbit pointers to next higher value
// using merge sort
#include
#include
/* Link list node */
struct Node
{
int data;
struct Node* next, *arbit;
};
/* function prototypes */
struct Node* SortedMerge(struct Node* a, struct Node* b);
void FrontBackSplit(struct Node* source,
struct Node** frontRef, struct Node** backRef);
/* sorts the linked list formed by arbit pointers
(does not change next pointer or data) */
void MergeSort(struct Node** headRef)
{
struct Node* head = *headRef;
struct Node* a, *b;
/* Base case -- length 0 or 1 */
if ((head == NULL) || (head->arbit == NULL))
return;
/* Split head into 'a' and 'b' sublists */
FrontBackSplit(head, &a, &b);
/* Recursively sort the sublists */
MergeSort(&a);
MergeSort(&b);
/* answer = merge the two sorted lists together */
*headRef = SortedMerge(a, b);
}
/* See https://www.geeksforgeeks.org/?p=3622 for details of this
function */
struct Node* SortedMerge(struct Node* a, struct Node* b)
{
struct Node* result = NULL;
/* Base cases */
if (a == NULL)
return (b);
else if (b==NULL)
return (a);
/* Pick either a or b, and recur */
if (a->data <= b->data)
{
result = a;
result->arbit = SortedMerge(a->arbit, b);
}
else
{
result = b;
result->arbit = SortedMerge(a, b->arbit);
}
return (result);
}
/* Split the nodes of the given list into front and back halves,
and return the two lists using the reference parameters.
If the length is odd, the extra node should go in the front list.
Uses the fast/slow pointer strategy. */
void FrontBackSplit(struct Node* source,
struct Node** frontRef, struct Node** backRef)
{
struct Node* fast, *slow;
if (source==NULL || source->arbit==NULL)
{
/* length < 2 cases */
*frontRef = source;
*backRef = NULL;
return;
}
slow = source, fast = source->arbit;
/* Advance 'fast' two nodes, and advance 'slow' one node */
while (fast != NULL)
{
fast = fast->arbit;
if (fast != NULL)
{
slow = slow->arbit;
fast = fast->arbit;
}
}
/* 'slow' is before the midpoint in the list, so split it in two
at that point. */
*frontRef = source;
*backRef = slow->arbit;
slow->arbit = NULL;
}
/* Function to insert a node at the beginging 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 off the new node */
new_node->next = (*head_ref);
new_node->arbit = NULL;
/* move the head to point to the new node */
(*head_ref) = new_node;
}
// Utility function to print result linked list
void printListafter(struct Node *node, struct Node *anode)
{
printf("Traversal using Next Pointer\n");
while (node!=NULL)
{
printf("%d, ", node->data);
node = node->next;
}
printf("\nTraversal using Arbit Pointer\n");
while (anode!=NULL)
{
printf("%d, ", anode->data);
anode = anode->arbit;
}
}
// This function populates arbit pointer in every node to the
// next higher value. And returns pointer to the node with
// minimum value
struct Node* populateArbit(struct Node *head)
{
// Copy next pointers to arbit pointers
struct Node *temp = head;
while (temp != NULL)
{
temp->arbit = temp->next;
temp = temp->next;
}
// Do merge sort for arbitrary pointers
MergeSort(&head);
// Return head of arbitrary pointer linked list
return head;
}
/* Driver program to test above functions*/
int main()
{
/* Start with the empty list */
struct Node* head = NULL;
/* Let us create the list shown above */
push(&head, 3);
push(&head, 2);
push(&head, 10);
push(&head, 5);
/* Sort the above created Linked List */
struct Node *ahead = populateArbit(head);
printf("\nResult Linked List is: \n");
printListafter(head, ahead);
getchar();
return 0;
}
Java
// Java program to populate arbit pointers
// to next higher value using merge sort
class LinkedList
{
static Node head;
/* Link list node */
static class Node
{
int data;
Node next, arbit;
Node(int data)
{
this.data = data;
next = null;
arbit = null;
}
}
// Utility function to print result linked list
void printList(Node node, Node anode)
{
System.out.println("Traversal using Next Pointer");
while (node != null)
{
System.out.print(node.data + " ");
node = node.next;
}
System.out.println("\nTraversal using Arbit Pointer");
while (anode != null)
{
System.out.print(anode.data + " ");
anode = anode.arbit;
}
}
// This function populates arbit pointer in every node to the
// next higher value. And returns pointer to the node with
// minimum value
private Node populateArbit(Node start)
{
Node temp = start;
// Copy next pointers to arbit pointers
while (temp != null)
{
temp.arbit = temp.next;
temp = temp.next;
}
// Do merge sort for arbitrary pointers and
// return head of arbitrary pointer linked list
return MergeSort(start);
}
/* sorts the linked list formed by arbit pointers
(does not change next pointer or data) */
private Node MergeSort(Node start)
{
/* Base case -- length 0 or 1 */
if (start == null || start.arbit == null)
{
return start;
}
/* Split head into 'middle' and 'nextofmiddle' sublists */
Node middle = getMiddle(start);
Node nextofmiddle = middle.arbit;
middle.arbit = null;
/* Recursively sort the sublists */
Node left = MergeSort(start);
Node right = MergeSort(nextofmiddle);
/* answer = merge the two sorted lists together */
Node sortedlist = SortedMerge(left, right);
return sortedlist;
}
// Utility function to get the middle of the linked list
private Node getMiddle(Node source)
{
// Base case
if (source == null)
return source;
Node fastptr = source.arbit;
Node slowptr = source;
// Move fastptr by two and slow ptr by one
// Finally slowptr will point to middle node
while (fastptr != null)
{
fastptr = fastptr.arbit;
if (fastptr != null)
{
slowptr = slowptr.arbit;
fastptr = fastptr.arbit;
}
}
return slowptr;
}
private Node SortedMerge(Node a, Node b)
{
Node result = null;
/* Base cases */
if (a == null)
return b;
else if (b == null)
return a;
/* Pick either a or b, and recur */
if (a.data <= b.data)
{
result = a;
result.arbit = SortedMerge(a.arbit, b);
}
else
{
result = b;
result.arbit = SortedMerge(a, b.arbit);
}
return result;
}
// Driver code
public static void main(String[] args)
{
LinkedList list = new LinkedList();
/* Let us create the list shown above */
list.head = new Node(5);
list.head.next = new Node(10);
list.head.next.next = new Node(2);
list.head.next.next.next = new Node(3);
/* Sort the above created Linked List */
Node ahead = list.populateArbit(head);
System.out.println("Result Linked List is:");
list.printList(head, ahead);
}
}
// This code is contributed by shubham96301
Python3
# Python3 program to populate arbit pointers
# to next higher value using merge sort
head = None
# Link l node
class Node:
def __init__(self, data):
self.data = data
self.next = None
self.arbit = None
# Utility function to print result linked l
def printList(node, anode):
print("Traversal using Next Pointer")
while (node != None):
print(node.data, end = ", ")
node = node.next
print("\nTraversal using Arbit Pointer");
while (anode != None):
print(anode.data, end = ", ")
anode = anode.arbit
# This function populates arbit pointer in
# every node to the next higher value. And
# returns pointer to the node with minimum
# value
def populateArbit(start):
temp = start
# Copy next pointers to arbit pointers
while (temp != None):
temp.arbit = temp.next
temp = temp.next
# Do merge sort for arbitrary pointers and
# return head of arbitrary pointer linked l
return MergeSort(start)
# Sorts the linked l formed by arbit pointers
# (does not change next pointer or data)
def MergeSort(start):
# Base case -- length 0 or 1
if (start == None or start.arbit == None):
return start
# Split head into 'middle' and
# 'nextofmiddle' sublists
middle = getMiddle(start)
nextofmiddle = middle.arbit
middle.arbit = None
# Recursively sort the sublists
left = MergeSort(start)
right = MergeSort(nextofmiddle)
# answer = merge the two sorted lists together
sortedlist = SortedMerge(left, right)
return sortedlist
# Utility function to get the
# middle of the linked l
def getMiddle(source):
# Base case
if (source == None):
return source
fastptr = source.arbit
slowptr = source
# Move fastptr by two and slow ptr by one
# Finally slowptr will point to middle node
while (fastptr != None):
fastptr = fastptr.arbit
if (fastptr != None):
slowptr = slowptr.arbit
fastptr = fastptr.arbit
return slowptr
def SortedMerge(a, b):
result = None
# Base cases
if (a == None):
return b
elif (b == None):
return a
# Pick either a or b, and recur
if (a.data <= b.data):
result = a
result.arbit = SortedMerge(a.arbit, b)
else:
result = b
result.arbit = SortedMerge(a, b.arbit)
return result
# Driver code
if __name__=='__main__':
# Let us create the l shown above
head = Node(5)
head.next = Node(10)
head.next.next = Node(2)
head.next.next.next = Node(3)
# Sort the above created Linked List
ahead = populateArbit(head)
print("Result Linked List is:")
printList(head, ahead)
# This code is contributed by rutvik_56
C#
// C# program to populate arbit pointers
// to next higher value using merge sort
using System;
public class LinkedList
{
public Node head;
/* Link list node */
public class Node
{
public int data;
public Node next, arbit;
public Node(int data)
{
this.data = data;
next = null;
arbit = null;
}
}
// Utility function to print result linked list
void printList(Node node, Node anode)
{
Console.WriteLine("Traversal using Next Pointer");
while (node != null)
{
Console.Write(node.data + " ");
node = node.next;
}
Console.WriteLine("\nTraversal using Arbit Pointer");
while (anode != null)
{
Console.Write(anode.data + " ");
anode = anode.arbit;
}
}
// This function populates arbit pointer in every node to the
// next higher value. And returns pointer to the node with
// minimum value
private Node populateArbit(Node start)
{
Node temp = start;
// Copy next pointers to arbit pointers
while (temp != null)
{
temp.arbit = temp.next;
temp = temp.next;
}
// Do merge sort for arbitrary pointers and
// return head of arbitrary pointer linked list
return MergeSort(start);
}
/* sorts the linked list formed by arbit pointers
(does not change next pointer or data) */
private Node MergeSort(Node start)
{
/* Base case -- length 0 or 1 */
if (start == null || start.arbit == null)
{
return start;
}
/* Split head into 'middle' and 'nextofmiddle' sublists */
Node middle = getMiddle(start);
Node nextofmiddle = middle.arbit;
middle.arbit = null;
/* Recursively sort the sublists */
Node left = MergeSort(start);
Node right = MergeSort(nextofmiddle);
/* answer = merge the two sorted lists together */
Node sortedlist = SortedMerge(left, right);
return sortedlist;
}
// Utility function to get the middle of the linked list
private Node getMiddle(Node source)
{
// Base case
if (source == null)
return source;
Node fastptr = source.arbit;
Node slowptr = source;
// Move fastptr by two and slow ptr by one
// Finally slowptr will point to middle node
while (fastptr != null)
{
fastptr = fastptr.arbit;
if (fastptr != null)
{
slowptr = slowptr.arbit;
fastptr = fastptr.arbit;
}
}
return slowptr;
}
private Node SortedMerge(Node a, Node b)
{
Node result = null;
/* Base cases */
if (a == null)
return b;
else if (b == null)
return a;
/* Pick either a or b, and recur */
if (a.data <= b.data)
{
result = a;
result.arbit = SortedMerge(a.arbit, b);
}
else
{
result = b;
result.arbit = SortedMerge(a, b.arbit);
}
return result;
}
// Driver code
public static void Main(String[] args)
{
LinkedList list = new LinkedList();
/* Let us create the list shown above */
list.head = new Node(5);
list.head.next = new Node(10);
list.head.next.next = new Node(2);
list.head.next.next.next = new Node(3);
/* Sort the above created Linked List */
Node ahead = list.populateArbit(list.head);
Console.WriteLine("Result Linked List is:");
list.printList(list.head, ahead);
}
}
/* This code contributed by PrinciRaj1992 */
Javascript
输出:
Result Linked List is:
Traversal using Next Pointer
5, 10, 2, 3,
Traversal using Arbit Pointer
2, 3, 5, 10,
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