展平链表
给定一个链表，其中每个节点代表一个链表并包含两个其类型的指针：
(i) 指向主列表中下一个节点的指针（我们在下面的代码中称之为“右”指针）
(ii) 指向该节点所在的链表的指针（我们在下面的代码中将其称为“向下”指针）。
所有链表都已排序。看下面的例子
5 -> 10 -> 19 -> 28
| | | |
V V V V
7 20 22 35
| | |
V V V
8 50 40
| |
V V
30 45编写一个函数flatten() 将列表扁平化为单个链表。扁平化的链表也应该排序。例如，对于上面的输入列表，输出列表应该是 5->7->8->10->19->20->22->28->30->35->40->45->50 .
这个想法是对链表使用合并排序的 Merge() 过程。我们使用merge() 将列表一一合并。我们递归地将当前列表与已经展平的列表合并（）。
向下指针用于链接扁平列表的节点。
下面是上述方法的实现：
C++
// C++ program for flattening a Linked List
#include
using namespace std;
// Link list node
class Node
{
public:
int data;
Node *right, *down;
};
Node* head = NULL;
// An utility function to merge two sorted
// linked lists
Node* merge(Node* a, Node* b)
{
// If first linked list is empty then second
// is the answer
if (a == NULL)
return b;
// If second linked list is empty then first
// is the result
if (b == NULL)
return a;
// Compare the data members of the two linked
// lists and put the larger one in the result
Node* result;
if (a->data < b->data)
{
result = a;
result->down = merge(a->down, b);
}
else
{
result = b;
result->down = merge(a, b->down);
}
result->right = NULL;
return result;
}
Node* flatten(Node* root)
{
// Base Cases
if (root == NULL || root->right == NULL)
return root;
// Recur for list on right
root->right = flatten(root->right);
// Now merge
root = merge(root, root->right);
// Return the root
// it will be in turn merged with its left
return root;
}
// Utility function to insert a node at
// beginning of the linked list
Node* push(Node* head_ref, int data)
{
// Allocate the Node & Put in the data
Node* new_node = new Node();
new_node->data = data;
new_node->right = NULL;
// Make next of new Node as head
new_node->down = head_ref;
// Move the head to point to new Node
head_ref = new_node;
return head_ref;
}
void printList()
{
Node* temp = head;
while (temp != NULL)
{
cout << temp->data << " ";
temp = temp->down;
}
cout << endl;
}
// Driver code
int main()
{
/* Let us create the following linked list
5 -> 10 -> 19 -> 28
| | | |
V V V V
7 20 22 35
| | |
V V V
8 50 40
| |
V V
30 45
*/
head = push(head, 30);
head = push(head, 8);
head = push(head, 7);
head = push(head, 5);
head->right = push(head->right, 20);
head->right = push(head->right, 10);
head->right->right = push(head->right->right, 50);
head->right->right = push(head->right->right, 22);
head->right->right = push(head->right->right, 19);
head->right->right->right = push(head->right->right->right, 45);
head->right->right->right = push(head->right->right->right, 40);
head->right->right->right = push(head->right->right->right, 35);
head->right->right->right = push(head->right->right->right, 20);
// Flatten the list
head = flatten(head);
printList();
return 0;
}
// This code is contributed by rajsanghavi9. Java
// Java program for flattening a Linked List
class LinkedList
{
Node head; // head of list
/* Linked list Node*/
class Node
{
int data;
Node right, down;
Node(int data)
{
this.data = data;
right = null;
down = null;
}
}
// An utility function to merge two sorted linked lists
Node merge(Node a, Node b)
{
// if first linked list is empty then second
// is the answer
if (a == null) return b;
// if second linked list is empty then first
// is the result
if (b == null) return a;
// compare the data members of the two linked lists
// and put the larger one in the result
Node result;
if (a.data < b.data)
{
result = a;
result.down = merge(a.down, b);
}
else
{
result = b;
result.down = merge(a, b.down);
}
result.right = null;
return result;
}
Node flatten(Node root)
{
// Base Cases
if (root == null || root.right == null)
return root;
// recur for list on right
root.right = flatten(root.right);
// now merge
root = merge(root, root.right);
// return the root
// it will be in turn merged with its left
return root;
}
/* Utility function to insert a node at beginning of the
linked list */
Node push(Node head_ref, int data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(data);
/* 3. Make next of new Node as head */
new_node.down = head_ref;
/* 4. Move the head to point to new Node */
head_ref = new_node;
/*5. return to link it back */
return head_ref;
}
void printList()
{
Node temp = head;
while (temp != null)
{
System.out.print(temp.data + " ");
temp = temp.down;
}
System.out.println();
}
/* Driver program to test above functions */
public static void main(String args[])
{
LinkedList L = new LinkedList();
/* Let us create the following linked list
5 -> 10 -> 19 -> 28
| | | |
V V V V
7 20 22 35
| | |
V V V
8 50 40
| |
V V
30 45
*/
L.head = L.push(L.head, 30);
L.head = L.push(L.head, 8);
L.head = L.push(L.head, 7);
L.head = L.push(L.head, 5);
L.head.right = L.push(L.head.right, 20);
L.head.right = L.push(L.head.right, 10);
L.head.right.right = L.push(L.head.right.right, 50);
L.head.right.right = L.push(L.head.right.right, 22);
L.head.right.right = L.push(L.head.right.right, 19);
L.head.right.right.right = L.push(L.head.right.right.right, 45);
L.head.right.right.right = L.push(L.head.right.right.right, 40);
L.head.right.right.right = L.push(L.head.right.right.right, 35);
L.head.right.right.right = L.push(L.head.right.right.right, 20);
// flatten the list
L.head = L.flatten(L.head);
L.printList();
}
} /* This code is contributed by Rajat Mishra */Python3
# Python program for flattening a Linked List
class Node():
def __init__(self,data):
self.data = data
self.right = None
self.down = None
class LinkedList():
def __init__(self):
# head of list
self.head = None
# Utility function to insert a node at beginning of the
# linked list
def push(self,head_ref,data):
# 1 & 2: Allocate the Node &
# Put in the data
new_node = Node(data)
# Make next of new Node as head
new_node.down = head_ref
# 4. Move the head to point to new Node
head_ref = new_node
# 5. return to link it back
return head_ref
def printList(self):
temp = self.head
while(temp != None):
print(temp.data,end=" ")
temp = temp.down
print()
# An utility function to merge two sorted linked lists
def merge(self, a, b):
# if first linked list is empty then second
# is the answer
if(a == None):
return b
# if second linked list is empty then first
# is the result
if(b == None):
return a
# compare the data members of the two linked lists
# and put the larger one in the result
result = None
if (a.data < b.data):
result = a
result.down = self.merge(a.down,b)
else:
result = b
result.down = self.merge(a,b.down)
result.right = None
return result
def flatten(self, root):
# Base Case
if(root == None or root.right == None):
return root
# recur for list on right
root.right = self.flatten(root.right)
# now merge
root = self.merge(root, root.right)
# return the root
# it will be in turn merged with its left
return root
# Driver program to test above functions
L = LinkedList()
'''
Let us create the following linked list
5 -> 10 -> 19 -> 28
| | | |
V V V V
7 20 22 35
| | |
V V V
8 50 40
| |
V V
30 45
'''
L.head = L.push(L.head, 30);
L.head = L.push(L.head, 8);
L.head = L.push(L.head, 7);
L.head = L.push(L.head, 5);
L.head.right = L.push(L.head.right, 20);
L.head.right = L.push(L.head.right, 10);
L.head.right.right = L.push(L.head.right.right, 50);
L.head.right.right = L.push(L.head.right.right, 22);
L.head.right.right = L.push(L.head.right.right, 19);
L.head.right.right.right = L.push(L.head.right.right.right, 45);
L.head.right.right.right = L.push(L.head.right.right.right, 40);
L.head.right.right.right = L.push(L.head.right.right.right, 35);
L.head.right.right.right = L.push(L.head.right.right.right, 20);
# flatten the list
L.head = L.flatten(L.head);
L.printList()
# This code is contributed by maheshwaripiyush9C#
// C# program for flattening a Linked List
using System;
public class List {
Node head; // head of list
/* Linked list Node */
public
class Node {
public
int data;
public
Node right, down;
public
Node(int data) {
this.data = data;
right = null;
down = null;
}
}
// An utility function to merge two sorted linked lists
Node merge(Node a, Node b) {
// if first linked list is empty then second
// is the answer
if (a == null)
return b;
// if second linked list is empty then first
// is the result
if (b == null)
return a;
// compare the data members of the two linked lists
// and put the larger one in the result
Node result;
if (a.data < b.data) {
result = a;
result.down = merge(a.down, b);
}
else {
result = b;
result.down = merge(a, b.down);
}
result.right = null;
return result;
}
Node flatten(Node root) {
// Base Cases
if (root == null || root.right == null)
return root;
// recur for list on right
root.right = flatten(root.right);
// now merge
root = merge(root, root.right);
// return the root
// it will be in turn merged with its left
return root;
}
/*
* Utility function to insert a node at beginning of the linked list
*/
Node Push(Node head_ref, int data) {
/*
* 1 & 2: Allocate the Node & Put in the data
*/
Node new_node = new Node(data);
/* 3. Make next of new Node as head */
new_node.down = head_ref;
/* 4. Move the head to point to new Node */
head_ref = new_node;
/* 5. return to link it back */
return head_ref;
}
void printList() {
Node temp = head;
while (temp != null) {
Console.Write(temp.data + " ");
temp = temp.down;
}
Console.WriteLine();
}
/* Driver program to test above functions */
public static void Main(String []args) {
List L = new List();
/*
* Let us create the following linked list 5 -> 10 -> 19 -> 28 | | | | V V V V 7
* 20 22 35 | | | V V V 8 50 40 | | V V 30 45
*/
L.head = L.Push(L.head, 30);
L.head = L.Push(L.head, 8);
L.head = L.Push(L.head, 7);
L.head = L.Push(L.head, 5);
L.head.right = L.Push(L.head.right, 20);
L.head.right = L.Push(L.head.right, 10);
L.head.right.right = L.Push(L.head.right.right, 50);
L.head.right.right = L.Push(L.head.right.right, 22);
L.head.right.right = L.Push(L.head.right.right, 19);
L.head.right.right.right = L.Push(L.head.right.right.right, 45);
L.head.right.right.right = L.Push(L.head.right.right.right, 40);
L.head.right.right.right = L.Push(L.head.right.right.right, 35);
L.head.right.right.right = L.Push(L.head.right.right.right, 20);
// flatten the list
L.head = L.flatten(L.head);
L.printList();
}
}
// This code is contributed by umadevi9616Javascript
输出：
5 7 8 10 19 20 20 22 30 35 40 45 50如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。