python实现链表
# @File: linklist class Node(object): def __init__(self, data): self.data = data self.next = None # 链表的实现:带头节点的链表 不带头节点的链表 # 带头节点的链表 # 头插法 def create_link_list_head(li): head = Node(None) for num in li: p = Node(num) p.next = head.next head.next = p return head # 尾插法 def create_link_list_tail(li): head = Node(None) tail = head for num in li: p = Node(num) tail.next = p tail = p return head # 打印输出 def traverse_link_list(head): p = head.next while p: print(p.data) p = p.next # head = create_link_list_head([1, 2, 3, 4, 5]) # head = create_link_list_tail([1, 2, 3, 4, 5]) # traverse_link_list(head) # 不带头节点的链表 # 头插法 def create_link_list_head2(li): head = None for num in li: p = Node(num) p.next = head head = p return head # 尾插法 def create_link_list_tail2(li): head = None tail = head for num in li: p = Node(num) if tail: tail.next = p tail = p else: tail = p head = p return head # 打印 def traverse_link_list2(head): p = head while p: print(p.data) p = p.next head = create_link_list_head2([1, 2, 3, 4, 5]) head = create_link_list_tail2([1, 2, 3, 4, 5]) traverse_link_list2(head)
# s = iter([1, 2, 3]) # print(s.__next__()) class Node(object): def __init__(self, data): self.data = data self.next = None # 头插法 (带头节点的方式) def create(li): head = Node(None) for num in li: p = Node(num) p.next = head.next head.next = p return head # 计算链表的长度 def size(head): count = 0 p = head.next while p: count += 1 p = p.next return count # 查询链表中元素 def search(head, item): current = head found = False while current and not found: if current.data == item: found = True else: current = current.next return found # 删除链表中元素 def remove(head, item): current = head previous = None found = False while not found: if current.data == item: found = True else: previous = current current = current.next previous.next = current.next # 打印 def print_list(head): p = head.next while p: print(p.data) p = p.next head = create([1, 2, 3, 4, 5]) print_list(head) # print_list(head) # print(size(head)) # print(search(head, 11)) print("remove", remove(head, 1)) print_list(head)
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