Python3程序将链表的子列表从位置M到N向右旋转K位

给定一个链表和两个位置“m”和“n”。任务是将子列表从位置 m 旋转到 n，向右旋转 k 个位置。

例子：

Input: list = 1->2->3->4->5->6, m = 2, n = 5, k = 2
Output: 1->4->5->2->3->6
Rotate the sublist 2 3 4 5 towards right 2 times
then the modified list are: 1 4 5 2 3 6

Input: list = 20->45->32->34->22->28, m = 3, n = 6, k = 3
Output: 20->45->34->22->28->32
Rotate the sublist 32 34 22 28 towards right 3 times
then the modified list are: 20 45 34 22 28 32

编程需要懂一点英语

方法：为了旋转从 m 到 n 元素的给定子列表，将列表从第 (n-k+1)^个节点移动到第 n^个节点到子列表的开头以完成旋转。
如果 k 大于子列表的大小，那么我们将取其与子列表大小的模数。因此，使用指针和计数器遍历列表，我们将保存第 (m-1)^个节点，然后使其指向第 (n-k+1)^个节点，从而将第 (n-k+1)^个节点带到子列表的开始（前）。
同样，我们将保存第 m^个节点，然后让第 n^个节点指向它。为了保持列表的其余部分完好无损，我们将使 (nk) ^th节点指向 n 的下一个节点（可能为 NULL）。最后我们将得到 k 次右旋子列表。

下面是上述方法的实现：

Python3

# Python3 implementation of the above approach
import math
  
# Definition of node of linkedlist
class Node: 
    def __init__(self, data): 
        self.data = data 
        self.next = None
  
# This function take head pointer of list, 
# start and end points of sublist that is 
# to be rotated and the number k and
# rotate the sublist to right by k places.
def rotateSubList(A, m, n, k):
    size = n - m + 1
  
    # If k is greater than size of sublist then 
    # we will take its modulo with size of sublist
    if (k > size):
        k = k % size
      
    # If k is zero or k is equal to size or k is
    # a multiple of size of sublist then list 
    # remains intact
    if (k == 0 or k == size):
        head = A
        while (head != None):
            print(head.data)
            head = head.next
          
        return
      
    link = None # m-th node
    if (m == 1) :
        link = A
      
    # This loop will traverse all node till
    # end node of sublist. 
    c = A # Current traversed node
    count = 0 # Count of traversed nodes
    end = None
    pre = None # Previous of m-th node
    while (c != None) :
        count = count + 1
  
        # We will save (m-1)th node and later
        # make it point to (n-k+1)th node
        if (count == m - 1) :
            pre = c
            link = c.next
          
        if (count == n - k) :
            if (m == 1) :
                end = c
                A = c.next
              
            else :
                end = c
  
                # That is how we bring (n-k+1)th
                # node to front of sublist.
                pre.next = c.next
              
        # This keeps rest part of list intact.
        if (count == n) :
            d = c.next
            c.next = link
            end.next = d
            head = A
            while (head != None) :
                print(head.data, end = " ")
                head = head.next
              
            return
          
        c = c.next
      
# Function for creating and linking new nodes
def push(head, val):
    new_node = Node(val)
    new_node.data = val
    new_node.next = head
    head = new_node
    return head
  
# Driver code
if __name__=='__main__': 
    head = None
    head = push(head, 70)
    head = push(head, 60)
    head = push(head, 50)
    head = push(head, 40)
    head = push(head, 30)
    head = push(head, 20)
    head = push(head, 10)
    tmp = head
    print("Given List: ", end = "")
    while (tmp != None) :
        print(tmp.data, end = " ")
        tmp = tmp.next
      
    print()
  
    m = 3
    n = 6
    k = 2 
    print("After rotation of sublist: ", end = "")
    rotateSubList(head, m, n, k)
  
# This code is contributed by Srathore

输出：

Given List: 10 20 30 40 50 60 70 
After rotation of sublist: 10 20 50 60 30 40 70

请参阅完整文章将链表的子列表从位置 M 向右旋转 K 位以获取更多详细信息！