用于就地重新排列给定链表的Python程序

给定一个单链表 L ₀ -> L ₁ -> ... -> L _n-1 -> L _n 。重新排列列表中的节点，使新形成的列表为： L ₀ -> L _n -> L ₁ -> L _n-1 -> L ₂ -> L _n-2 ...
您需要在不更改节点值的情况下就地执行此操作。

例子：

Input: 1 -> 2 -> 3 -> 4
Output: 1 -> 4 -> 2 -> 3

Input: 1 -> 2 -> 3 -> 4 -> 5
Output: 1 -> 5 -> 2 -> 4 -> 3

简单的解决方案：

1) Initialize current node as head.
2) While next of current node is not null, do following
    a) Find the last node, remove it from the end and insert it as next
       of the current node.
    b) Move current to next to next of current

上述简单解决方案的时间复杂度为 O(n ² )，其中 n 是链表中的节点数。

更好的解决方案：
1) 将给定链表的内容复制到向量。
2）通过交换两端的节点来重新排列给定的向量。
3) 将修改后的向量复制回链表。
这种方法的实施：https://ide.geeksforgeeks.org/1eGSEy
感谢 Arushi Dhamija 提出这种方法。

高效解决方案：

1) Find the middle point using tortoise and hare method.
2) Split the linked list into two halves using found middle point in step 1.
3) Reverse the second half.
4) Do alternate merge of first and second halves.

该解决方案的时间复杂度为 O(n)。

下面是这个方法的实现。

Python3

# Python program to rearrange linked list 
# in place
  
# Node Class
class Node:
    
    # Constructor to create 
    # a new node
    def __init__(self, d):
        self.data = d
        self.next = None
          
def printlist(node):
    if(node == None):
        return
    while(node != None):
        print(node.data," -> ", 
              end = "")
        node = node.next
  
def reverselist(node):
    prev = None
    curr = node
    next=None
    while (curr != None):
        next = curr.next
        curr.next = prev
        prev = curr
        curr = next
    node = prev
    return node
  
def rearrange(node):
    
    # 1) Find the middle point using 
    # tortoise and hare method
    slow = node
    fast = slow.next
  
    while (fast != None and 
           fast.next != None):
        slow = slow.next
        fast = fast.next.next
      
    # 2) Split the linked list in 
    # two halves
    # node1, head of first half    
    # 1 -> 2 -> 3
    # node2, head of second half   
    # 4 -> 5    
    node1 = node
    node2 = slow.next
    slow.next = None
      
    # 3) Reverse the second half, 
    # i.e., 5 -> 4
    node2 = reverselist(node2)
      
    # 4) Merge alternate nodes
    # Assign dummy Node
    node = Node(0)  
      
    # curr is the pointer to this 
    # dummy Node, which will be 
    # used to form the new list
    curr = node
      
    while (node1 != None or 
           node2 != None):
          
        # First add the element from 
        # first list
        if (node1 != None):
            curr.next = node1
            curr = curr.next
            node1 = node1.next
          
        # Then add the element from 
        # second list
        if(node2 != None):
            curr.next = node2
            curr = curr.next
            node2 = node2.next
      
    # Assign the head of the new list 
    # to head pointer
    node = node.next
  
head = None
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
head.next.next.next = Node(4)
head.next.next.next.next = Node(5)
  
# Print original list
printlist(head) 
  
# Rearrange list as per ques
rearrange(head) 
print()
  
# Print modified list
printlist(head) 
# This code is contributed by ab2127

输出：

1 -> 2 -> 3 -> 4 -> 5 
1 -> 5 -> 2 -> 4 -> 3

时间复杂度： O(n)
辅助空间： O(1)
感谢 Gaurav Ahirwar 提出上述方法。

另一种方法：
1.取两个指针prev和curr，分别保存head和head->next的地址。
2.比较他们的数据并交换。
之后，形成一个新的链表。

下面是实现：

Python3

# Python3 code to rearrange linked list 
# in place
class Node:
  
    def __init__(self, x):
        self.data = x
        self.next = None
  
# Function for rearranging a 
# linked list with high and 
# low value
def rearrange(head):
  
    # Base case
    if (head == None):
        return head
  
    # Two pointer variable
    prev, curr = head, head.next
  
    while (curr):
  
        # Swap function for swapping 
        # data
        if (prev.data > curr.data):
            prev.data, curr.data = curr.data, prev.data
  
        # Swap function for swapping data
        if (curr.next and curr.next.data > curr.data):
            curr.next.data, curr.data = curr.data, curr.next.data
  
        prev = curr.next
  
        if (not curr.next):
            break
  
        curr = curr.next.next
  
    return head
  
# Function to insert a node in the
# linked list at the beginning
def push(head, k):
  
    tem = Node(k)
    tem.data = k
    tem.next = head
    head = tem
    return head
  
# Function to display node of 
# linked list
def display(head):
    curr = head
  
    while (curr != None):
        print(curr.data, end = " ")
        curr = curr.next
  
# Driver code
if __name__ == '__main__':
    head = None
  
    # Let create a linked list
    # 9 . 6 . 8 . 3 . 7
    head = push(head, 7)
    head = push(head, 3)
    head = push(head, 8)
    head = push(head, 6)
    head = push(head, 9)
    head = rearrange(head)
    display(head)
# This code is contributed by mohit kumar 29

输出：

6 9 3 8 7

时间复杂度： O(n)
辅助空间： O(1)
感谢 Aditya 提出这种方法。

另一种方法：（使用递归）

1. 持有指向头节点的指针并使用递归直到最后一个节点
2. 到达最后一个节点后，开始将最后一个节点交换到头节点的下一个节点
3. 将头指针移动到下一个节点
4. 重复此操作，直到头部和最后一个节点相遇或彼此相邻
5. 一旦满足停止条件，我们需要丢弃左节点以修复在交换节点时在列表中创建的循环。

Python3

# Python3 program to implement
# the above approach
class Node:    
    def __init__(self, key):        
        self.data = key
        self.next = None
  
left = None
  
# Function to print the list
def printlist(head):    
    while (head != None):
        print(head.data, end = " ")
        if (head.next != None):
            print("->", end = "")
              
        head = head.next
          
    print()
  
# Function to rearrange
def rearrange(head):    
    global left
    if (head != None):
        left = head
        reorderListUtil(left)
  
def reorderListUtil(right):    
    global left
    if (right == None):
        return
      
    reorderListUtil(right.next)
      
    # We set left = null, when we 
    # reach stop condition, so no 
    # processing required after that
    if (left == None):
        return
      
    # Stop condition: odd case : 
    # left = right, even
    # case : left.next = right
    if (left != right and 
        left.next != right):
        temp = left.next
        left.next = right
        right.next = temp
        left = temp
    else:
          
        # Stop condition , set null 
        # to left nodes
        if (left.next == right):
              
            # Even case
            left.next.next = None
            left = None
        else:
              
            # Odd case
            left.next = None
            left = None
  
# Driver code
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
head.next.next.next = Node(4)
head.next.next.next.next = Node(5)
  
# Print original list
printlist(head)
  
#  Modify the list
rearrange(head)
  
# Print modified list
printlist(head)
# This code is contributed by patel2127

输出：

1 ->2 ->3 ->4 ->5 
1 ->5 ->2 ->4 ->3

请参阅有关就地重新排列给定链接列表的完整文章。更多细节！