Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes, in the end, should remain as it is.

You may not alter the values in the list's nodes, only nodes themselves may be changed.

 

Example 1:

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

Example 2:

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

Example 3:

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

Example 4:

Input: head = [1], k = 1
Output: [1]

 

Constraints:

• The number of nodes in the list is in the range sz.
• 1 <= sz <= 5000
• 0 <= Node.val <= 1000
• 1 <= k <= sz

 

Follow-up: Can you solve the problem in O(1) extra memory space?

 

Code: get length of the list n, 得到times 和rem， 然后建一个reverse的funcion， reverse k 个node， 再返回，新list 的head， 新list的tail 和右半部分的list的head。

同时利用一个dummy来作为pre，再不断更新pre和head， 最后返回dummy.next.

 

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
        if k == 1: return head
        n = self.getLength(head)
        times, rem = divmod(n, k)
        dummy = ListNode()
        pre = dummy
        for _ in range(times):
            middleHead, middleTail, rightHead = self.reverse(head, k)
            pre.next = middleHead
            pre= middleTail
            head = rightHead
        middleTail.next = rightHead
        return dummy.next
    
    def getLength(self, head):
        count = 0
        while head:
            count += 1
            head = head.next
        return count
    
    def reverse(self, head, k) -> ('ListNode', 'ListNode', 'ListNode'):
        pre, oriHead = None, head
        for _ in range(k):
            temp = head.next
            head.next = pre
            pre= head
            head = temp
        return pre, oriHead, head