用于合并两个已排序链表的Python程序，使得合并后的链表顺序相反
给定两个按升序排序的链表。合并它们，使结果列表按降序（倒序）。
例子：
Input: a: 5->10->15->40
b: 2->3->20
Output: res: 40->20->15->10->5->3->2
Input: a: NULL
b: 2->3->20
Output: res: 20->3->2一个简单的解决方案是执行以下操作。
1）反转第一个列表'a'。
2）反转第二个列表'b'。
3）合并两个反向列表。
另一个简单的解决方案是首先合并两个列表，然后反转合并的列表。
以上两种方案都需要对链表进行两次遍历。
如何在没有反向、O(1) 辅助空间（就地）且仅遍历两个列表的情况下求解？
这个想法是遵循合并样式过程。将结果列表初始化为空。从头到尾遍历两个列表。比较两个列表的当前节点，并在结果列表的开头插入两个中较小的一个。
1) Initialize result list as empty: res = NULL.
2) Let 'a' and 'b' be heads first and second lists respectively.
3) While (a != NULL and b != NULL)
a) Find the smaller of two (Current 'a' and 'b')
b) Insert the smaller value node at the front of the result.
c) Move ahead in the list of the smaller nodes.
4) If 'b' becomes NULL before 'a', insert all nodes of 'a'
into the result list at the beginning.
5) If 'a' becomes NULL before 'b', insert all nodes of 'a'
into result list at the beginning. 以下是上述解决方案的实现。
Python3
# Given two sorted non-empty linked lists.
# Merge them in such a way that the result
# list will be in reverse order. Reversing
# of linked list is not allowed. Also,
# extra space should be O(1)
# Node of a linked list
class Node:
def __init__(self, next = None,
data = None):
self.next = next
self.data = data
# Given two non-empty linked lists
# 'a' and 'b'
def SortedMerge(a,b):
# If both lists are empty
if (a == None and b == None):
return None
# Initialize head of the
# resultant list
res = None
# Traverse both lists while both
# of then have nodes.
while (a != None and b != None):
# If a's current value is smaller
# or equal to b's current value.
if (a.key <= b.key):
# Store next of current Node
# in first list
temp = a.next
# Add 'a' at the front of
# resultant list
a.next = res
res = a
# Move ahead in first list
a = temp
# If a's value is greater. Below steps
# are similar to above (Only 'a' is
# replaced with 'b')
else:
temp = b.next
b.next = res
res = b
b = temp
# If second list reached end, but first
# list has nodes. Add remaining nodes of
# first list at the front of result list
while (a != None):
temp = a.next
a.next = res
res = a
a = temp
# If first list reached end, but second
# list has node. Add remaining nodes of
# first list at the front of result list
while (b != None):
temp = b.next
b.next = res
res = b
b = temp
return res
# Function to print Nodes in a given
# linked list
def printList(Node):
while (Node != None):
print( Node.key, end = " ")
Node = Node.next
# Utility function to create a new
# node with given key
def newNode(key):
temp = Node()
temp.key = key
temp.next = None
return temp
# Driver code
# Start with the empty list
res = None
# Let us create two sorted linked lists
# to test the above functions. Created
# lists shall be
# a: 5.10.15
# b: 2.3.20
a = newNode(5)
a.next = newNode(10)
a.next.next = newNode(15)
b = newNode(2)
b.next = newNode(3)
b.next.next = newNode(20)
print("List A before merge: ")
printList(a)
print("
List B before merge: ")
printList(b)
# Merge 2 increasing order LLs
# in descresing order
res = SortedMerge(a, b)
print("
Merged Linked List is: ")
printList(res)
# This code is contributed by Arnab Kundu输出：
List A before merge:
5 10 15
List B before merge:
2 3 20
Merged Linked List is:
20 15 10 5 3 2 时间复杂度： O(N)
辅助空间： O(1)
请参阅有关合并两个排序链表的完整文章，以便合并列表以相反的顺序获取更多详细信息！