Python刷题合集-算法编程题.zip

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 head == None or k == 1 or head.next == None: return head cur_node = head size = 0 while cur_node != None: size += 1 cur_node = cur_node.next j = size - size % k # 先计算要反转的个数 res = ListNode(-1) res.next = head cur_node, i = res, 0 while cur_node != None: if i == 0: anchor_node = cur_node cur_node = cur_node.next i += 1 continue if i > j: # 超过反转个数就停止循环 break if i % k == 0: # 第k个点 end_node.next = cur_node.next cur_node.next = anchor_node.next anchor_node.next = cur_node cur_node = end_node.next anchor_node = end_node # 完成一组反转，更新anchor_node elif i % k == 1: # 第1个点不动 end_node = cur_node cur_node = cur_node.next elif 1 < i % k < k: # 1到k之间的点插到链表段前面 end_node.next = cur_node.next cur_node.next = anchor_node.next anchor_node.next = cur_node cur_node = end_node.next i += 1 return res.next