python3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
        g= []
        res = []
        while head:
            g.append(head.val)
            head = head.next
        n = len(g)
        beishu = n//k
        for i in range(beishu):
            cur = g[i*k:i*k+k:1]
            # cur = sorted(cur,reverse=True)
            cur = cur[::-1]
            # res.append(cur)
            res += cur
        yushu = n%k
        if yushu:
            cur = g[beishu*k:n:1]
            # res.append(cur)
            res += cur
        hhh = ListNode(0)
        t = hhh
        for i in range(len(res)):
            t.next = ListNode(res[i])
            t = t.next
        return hhh.next