python链表——链表相连3个节点翻转

输入：1->2->3->4->5->6->7->8

输出：3->2->1->6->5->4->7->8

 

代码

def reverse_3(head):
    if head is None or head.next is None or head.next.next is None:
        return head
    cur1 = head
    cur2 = head.next
    cur3 = head.next.next
    head = cur3
    while cur3.next and cur3.next.next and cur3.next.next.next:
        # 赋值的指针操作
        nex1 = cur3.next
        nex2 = cur3.next.next
        nex3 = cur3.next.next.next
        # 3个节点内,指针转向
        cur3.next = cur2
        cur2.next = cur1
        cur1.next = nex3
        # 移动三个操作变量,到下一轮的3个节点
        cur1 = nex1
        cur2 = nex2
        cur3 = nex3
    # 循环结束时,将未完成的指针方向,补充上去
    nex1 = cur3.next
    cur3.next = cur2
    cur2.next = cur1
    cur1.next = nex1
    return head

# 测试
link = LinkedList()
print("=========before============")
link.add(1)
link.add(2)
link.add(3)
link.add(4)
link.add(5)
link.add(6)
link.add(7)
link.add(8)
link.print_link()
print("=========after============")
new_head = reverse_3(link.head)

while new_head:
    print(new_head.data)
    new_head = new_head.next

# 结果
=========before============
1
2
3
4
5
6
7
8
=========after============
3
2
1
6
5
4
7
8