扁平化链表的Java程序
给定一个链表，其中每个节点都表示一个链表并包含其类型的两个指针：
- 指向主列表中下一个节点的指针（我们在下面的代码中称其为“右”指针）。
- 指向该节点指向的链表的指针（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。请参阅以下示例
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() 将列表展平为单个链表。扁平化的链表也应该排序。比如上面的输入列表，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() 来一个一个地合并列表。我们递归地合并（）当前列表与已经展平的列表。
向下指针用于链接展平列表的节点。
下面是上述方法的实现：
Java
// Java program for flattening
// a Linked List
class LinkedList
{
// Head of list
Node head;
// 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 code
public static void main(String args[])
{
LinkedList L = new LinkedList();
/* 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输出：
5 7 8 10 19 20 20 22 30 35 40 45 50有关详细信息，请参阅有关扁平化链接列表的完整文章！