Leetcode 61: rotating linked list

Title Description

Give you a list of the head node head , Rotate the list , Move each node of the list to the right k A place .

Example 1

Input ：head = [1,2,3,4,5], k = 2
Output ：[4,5,1,2,3]

Example 2

Input ：head = [0,1,2], k = 4
Output ：[2,0,1]

Tips

• The number of nodes in the linked list is in the range [0, 500] Inside
• -100 <= Node.val <= 100
• 0 <= k <= 2 * $10^9$

Ideas

because k It could be worth a lot , So we have to put k except len Remainder .
The essence of this question is to find the penultimate k Nodes , Adjust all the nodes behind it to the front of the linked list .
Find the last k Nodes ： Define two nodes , Keep them apart k Nodes , Move back together .

Code

class Solution:
    def rotateRight(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        l = 0
        cur = head
        while cur is not None:
            l+=1
            cur = cur.next
        k = k%l
        bef = head
        aft = head
        for i in range(k):
            aft = aft.next
        while(aft.next is not None):
            aft = aft.next
            bef = bef.next

        # print(bef.val, aft.val)

        aft.next = head
        res = bef.next
        bef.next = None
        return res