给定一个 Linkedlist 和两个索引A和B ，任务是打印一个从A开始到B结束的子列表。
例子：
Input: list = 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 10 -> NULL, A = 3, B = 9
Output: 3 4 5 6 7 8 9
Input: list = 1 -> 3 -> 4 -> 10 -> NULL, A = 2, B = 2
Output: 3
方法：
请按照以下步骤解决问题：
- 创建三个实例变量：
- 当前：给定链表的头部
- subcurrent：子列表的头部
- subend：子列表的尾部。
- 遍历链表并初始化一个index_count变量，该变量在每次迭代后递增。
- 当index_count的值等于 A 时，将subcurrent设置为当前指向的节点。
- 如果index_count等于 B，则将 subend分配给正在引用的当前节点。将subend.next指向NULL以表示子列表的结尾。
- 将列表内容从subcurrent打印到subend 。
下面的代码是上述方法的实现：
Java
// Java Program to find the
// subList in a linked list
import java.util.Scanner;
// Class representing the
// structure of a Linked List
public class LinkedListSublist {
Node head;
class Node {
int data;
Node next = null;
Node(int new_data)
{
data = new_data;
}
}
// Function to push node
// at beginning of a
// Linked List
public void pushNode(int new_data)
{
Node new_node = new Node(new_data);
new_node.next = head;
head = new_node;
}
// Function to find sublist
public Node subList(Node head,
int A,
int B)
{
Node subcurrent = null;
Node subend = null;
Node current = head;
int i = 1;
// traverse between indices
while (current != null
&& i <= B) {
// If the starting index
// of the sublist is found
if (i == A) {
subcurrent = current;
}
// If the ending index of
// the sublist is found
if (i == B) {
subend = current;
subend.next = null;
}
// Move to next node
current = current.next;
i++;
}
// Return the head
// of the sublist
return subcurrent;
}
// Function to print
// the linked list
public void traversing()
{
Node current = head;
while (current != null) {
System.out.print(current.data
+ " -> ");
current = current.next;
}
}
// Driver Program
public static void main(String args[])
{
LinkedListSublist list
= new LinkedListSublist();
int N = 1;
int value = 10;
while (N < 11) {
list.pushNode(value--);
N++;
}
// Starting index
int A = 3;
// Ending index
int B = 9;
list.head
= list.subList(
list.head, A, B);
list.traversing();
}
}Python3
# Python3 program to find the
# subList in a linked list
class Node:
def __init__(self, data):
self.data = data
self.next = None
# Class representing the
# structure of a Linked List
class LinkedListSublist:
def __init__(self):
self.head = None
# Function to push node
# at beginning of a
# Linked List
def pushNode(self, new_data):
new_node = Node(new_data)
new_node.next = self.head
self.head = new_node
# Function to find sublist
def subList(self, head, A, B):
subcurrent = None
subend = None
current = self.head
i = 1
# Traverse between indices
while (current != None and i <= B):
# If the starting index
# of the sublist is found
if (i == A):
subcurrent = current
# If the ending index of
# the sublist is found
if (i == B):
subend = current
subend.next = None
# Move to next node
current = current.next
i += 1
# Return the head
# of the sublist
return subcurrent
# Function to print
# the linked list
def traversing(self):
current = self.head
while (current != None):
print(current.data, end = " -> ")
current = current.next
# Driver Code
if __name__=='__main__':
list = LinkedListSublist()
N = 1
value = 10
while (N < 11):
list.pushNode(value)
value -= 1
N += 1
# Starting index
A = 3
# Ending index
B = 9
list.head = list.subList(list.head, A, B)
list.traversing()
# This code is contributed by pratham76C#
// C# Program to find the
// subList in a linked list
using System;
// Class representing the
// structure of a Linked List
public class LinkedListSublist
{
public Node head;
public class Node
{
public int data;
public Node next = null;
public Node(int new_data)
{
data = new_data;
}
}
// Function to push node
// at beginning of a
// Linked List
public void pushNode(int new_data)
{
Node new_node = new Node(new_data);
new_node.next = head;
head = new_node;
}
// Function to find sublist
public Node subList(Node head,
int A,
int B)
{
Node subcurrent = null;
Node subend = null;
Node current = head;
int i = 1;
// traverse between indices
while (current != null
&& i <= B)
{
// If the starting index
// of the sublist is found
if (i == A)
{
subcurrent = current;
}
// If the ending index of
// the sublist is found
if (i == B)
{
subend = current;
subend.next = null;
}
// Move to next node
current = current.next;
i++;
}
// Return the head
// of the sublist
return subcurrent;
}
// Function to print
// the linked list
public void traversing()
{
Node current = head;
while (current != null)
{
Console.Write(current.data
+ " -> ");
current = current.next;
}
}
// Driver Program
public static void Main(string []args)
{
LinkedListSublist list
= new LinkedListSublist();
int N = 1;
int value = 10;
while (N < 11)
{
list.pushNode(value--);
N++;
}
// Starting index
int A = 3;
// Ending index
int B = 9;
list.head
= list.subList(
list.head, A, B);
list.traversing();
}
}
// This code is contributed by rutvik_56输出：
3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 ->时间复杂度： O(N)
辅助空间： O(1)
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