leetcodeday61–旋转链表

给你一个链表的头节点 head ，旋转链表，将链表每个节点向右移动 k个位置。

示例 1：

输入：head = [1,2,3,4,5], k = 2
输出：[4,5,1,2,3]

示例 2：

输入：head = [0,1,2], k = 4
输出：[2,0,1]

提示：

• 链表中节点的数目在范围 [0, 500] 内
• -100 <= Node.val <= 100
• 0 <= k <= 2 * 109

代码：

# @lc app=leetcode.cn id=61 lang=python3
#
# [61] 旋转链表
#

# @lc code=start
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def rotateRight(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        def lenlink(head):
            i=0
            while head != None:
                head=head.next
                i=i+1
            return i
        lens=lenlink(head)
        if lens==0:
            return head
        k=k%(lens) 
        newlist=ListNode(0,None)
        newcur=newlist
        cur1 = head
        cur2 = head
        for i in range(lens-k):
            cur1=cur1.next
            #print(cur1.val)
        for i in range(lens):
            if i < k:
               newcur.val= cur1.val
               print(newcur.val)
               cur1=cur1.next
               newcur.next=ListNode(0,None)
               newcur=newcur.next
            
            else:
               newcur.val= cur2.val
               #print(cur2.val)
               cur2=cur2.next
               if i<lens-1:
                    newcur.next=ListNode(0,None)
               newcur=newcur.next 
        return newlist