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:
        # write code here
        # 加个表头
        res = ListNode(-1)
        res.next = head

        pre = res
        pre.next = head
        cur = head

        m = 0
        while cur:
            m+=1
            cur = cur.next
        n = int(m/k)
        # print(n)
        cur = head
        for i in range(n):
            for j in range(k-1):
                temp = cur.next
                cur.next = temp.next
                temp.next = pre.next
                pre.next = temp
            if i == 0:
                res.next = pre.next
            pre = cur
            cur = cur.next
        return res.next