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
        if not head or not head.next or k == 1:
            return head
        count = 1
        pre, cur, nxt = None, head, head.next
        stack = []
        while nxt:
            stack.append(cur)
            if count % k == 0:
                tmp_head, tail = self.reverse(stack)
                if not pre:
                    head = tmp_head
                else:
                    pre.next = tmp_head
                pre, cur, nxt = tail, tail.next, nxt.next
            else:
                cur, nxt = cur.next, nxt.next
            count += 1

        if count % k == 0:
            stack.append(cur)
            tmp_head, _ = self.reverse(stack)
            if not pre:
                head = tmp_head
            else:
                pre.next = tmp_head
        
        return head
    
    def reverse(self, stack: list[ListNode]):
        cur = stack.pop()
        head = cur
        while stack:
            nxt = cur.next
            tmp = stack.pop()
            cur.next = tmp
            tmp.next = nxt
            cur = cur.next
        return head, cur