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
        newhead = ListNode(0)
        newhead.next = head
        p = newhead
        cur = p.next
        i = 0
        while head:
            i+=1
            head = head.next
        len = i   #前面这一段while为获取链表长度

        sum = k   #进行累计每做一组就+k直到超过原本长度，为什么一开始就要累计k，如果k大于len，那就不用翻转
        while sum <= len:
            for j in range(k-1): #一组内进行翻转
                pre = cur.next
                cur.next = pre.next
                pre.next = p.next
                p.next = pre
            sum += k #翻转一组后累计
            p = cur #翻转的头固定节点转换 参考BM2
            cur = p.next
        return newhead.next