题解 | 链表中的节点每k个一组翻转

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param head ListNode类 
# @param k int整型 
# @return ListNode类
#
class Solution:
    def reverseKGroup(self , head: ListNode, k: int) -> ListNode:
        dummy = ListNode(0)
        dummy.next = head
        if not head:
            return dummy.next

        node = dummy
        n = 0
        while node.next:
            n+=1
            node = node.next
        
        prev = dummy
        cur = head
        rounds = n//k
        
        print (n,k)
        for _ in range(rounds):
            for _ in range(k-1):
                temp = cur.next
                cur.next = temp.next
                temp.next = prev.next
                prev.next = temp
            prev = cur
            cur = cur.next
        return dummy.next


        # write code here