Python程序将一个链表合并到另一个链表的交替位置
给定两个链表,将第二个链表的节点插入第一个链表的第一个链表的交替位置。
例如,如果第一个列表是 5->7->17->13->11,第二个是 12->10->2->4->6,那么第一个列表应该变成 5->12->7- >10->17->2->13->4->11->6 和第二个列表应该为空。只有在有可用位置时才应插入第二个列表的节点。例如,如果第一个列表是 1->2->3,第二个列表是 4->5->6->7->8,那么第一个列表应该变成 1->4->2->5 ->3->6 和第二个列表到 7->8。
不允许使用额外的空间(不允许创建额外的节点),即必须就地完成插入。预期的时间复杂度为 O(n),其中 n 是第一个列表中的节点数。
这个想法是在第一个循环中有可用位置时运行一个循环,并通过更改指针插入第二个列表的节点。以下是这种方法的实现。
Python3
# Python program to merge a linked list
# into another at alternate positions
class Node(object):
def __init__(self, data:int):
self.data = data
self.next = None
class LinkedList(object):
def __init__(self):
self.head = None
def push(self, new_data:int):
new_node = Node(new_data)
new_node.next = self.head
# 4. Move the head to point to
# new Node
self.head = new_node
# Function to print linked list from
# the Head
def printList(self):
temp = self.head
while temp != None:
print(temp.data)
temp = temp.next
# Main function that inserts nodes of linked
# list q into p at alternate positions.
# Since head of first list never changes
# but head of second list/ may change,
# we need single pointer for first list and
# double pointer for second list.
def merge(self, p, q):
p_curr = p.head
q_curr = q.head
# swap their positions until one
# finishes off
while p_curr != None and q_curr != None:
# Save next pointers
p_next = p_curr.next
q_next = q_curr.next
# make q_curr as next of p_curr
# change next pointer of q_curr
q_curr.next = p_next
# change next pointer of p_curr
p_curr.next = q_curr
# update current pointers for next
# iteration
p_curr = p_next
q_curr = q_next
q.head = q_curr
# Driver code
llist1 = LinkedList()
llist2 = LinkedList()
# Creating Linked lists
# 1.
llist1.push(3)
llist1.push(2)
llist1.push(1)
llist1.push(0)
# 2.
for i in range(8, 3, -1):
llist2.push(i)
print("First Linked List:")
llist1.printList()
print("Second Linked List:")
llist2.printList()
# Merging the LLs
llist1.merge(p=llist1, q=llist2)
print("Modified first linked list:")
llist1.printList()
print("Modified second linked list:")
llist2.printList()
# This code is contributed by Deepanshu Mehta
输出:
First Linked List:
1 2 3
Second Linked List:
4 5 6 7 8
Modified First Linked List:
1 4 2 5 3 6
Modified Second Linked List:
7 8
时间复杂度: O(N)
辅助空间: O(1)
有关详细信息,请参阅有关将一个链表合并到另一个链表中的完整文章!