如何在 python 中将数据插入到单链表的中间？

Question

Node类和SingleLinkedList类代码：

class Node():
'''represents a node as a building block of a single linked list'''
def __init__(self, element, next_node=None):
    '''(Node, obj, Node) -> NoneType
    construct a node as building block of a single linked list'''

    self._element = element
    self._next = next_node

def set_next(self, next_node):
    '''(Node, Node) -> NoneType
    set node to point to next_node'''
    self._next = next_node

def set_element(self, element):
    '''(Node, obj) ->NoneType
    set the _element to a new value'''
    self._element = element

def get_next(self):
    '''(Node) -> Node
    returns the reference to next node'''
    return self._next

def get_element(self):
    '''(Node) -> obj
    returns the element of this node'''
    return self._element

def __str__(self):
    '''(Node) -> str
    returns the element of this node and the reference to next node'''
    return "(" + str(self._element) + ", " + str(hex(id(self._next))) + ")"

class SingleLinkedList():
''' represents a single linked list'''
def __init__(self):
    '''(SingleLinkedList) ->NoneType
    initializes the references of an empty SLL'''
    self._size = 0
    self._head = None
    self._tail = None
    self._value = None
    self._next = None
def set_head(self, new_head):
    '''(SingleLinkedList) -> None
    updates the head'''
    self._head = new_head
def set_tail(self, new_tail):
    '''(SingleLinkedList) -> None
    updates the tail'''
    self._tail = new_tail
def get_head(self):
    '''(SingleLinkedList) -> Node
    Return the pointer to the head'''
    return self._head 
def get_tail(self):
    '''(SingleLinkedList) -> Node
    Return the pointer to the tail'''
    return self._tail

def is_empty(self):
    '''(SingleLinkedList) -> bool
    returns true if no item is in this SLL'''
    return self._size == 0

def size(self):
    '''(SingleLinkedList) -> int
    returns the number of items in this SLL'''
    return self._size

def add_first(self, element):
    '''(SingleLinkedList, obj) -> NoneType
    adds a node to the first of the SLL'''
    # create a node that point to the head
    node = Node(element, self._head)
    # let head point to the node
    self._head = node
    # if this node is the first node in this SLL, tail should point to this node too
    if (self._size == 0):
        self._tail = node
    # increment the size
    self._size += 1

def add_last(self, element):
    '''(SingleLinkedList, obj) -> NoneType
    adds a node to the end of this SLL'''
    # create a node with the given element that points to None
    node = Node(element, None)
    # let the _next part of the tail to point to newly created node
    self._tail.set_next(node)
    #let tail to point to the added node
    self._tail = node
    # if this node is the first node in this SLL, let head to point to this node too
    if (self._size == 0):
        self._head = node
    # increment the size
    self._size += 1

def remove_first(self):
    '''(SingleLinkedList, obj) -> obj
    remvoe the node from the head of this SLL and returns the element stored in this node'''
    # sset element to None in case SLL was empty
    element = None
    # if SLL is not empty
    if (self._head is not None):
        # get the first node
        node = self._head
        # let head point to the second node
        self._head = self._head.get_next()
        # decrement the size
        self._size -= 1
        #set the _next of previous head to point to None (for garbage collection purpose)
        node.set_next(None)
        # get the element stored in the node
        element = node.get_element()
    #return the element of the removed node
    return element

def __str__(self):
    '''(SingleLinkedList) -> str
    returns the items in the SLL in a string form
    '''
    # define a node, which points to the head
    cur = self._head
    # define an empty string to be used as a container for the items in the SLL
    result = ""
    # loop over the SLL until you get to the end of the SLL
    while (cur is not None):
        # get the element that of the current node and attach it to the final result
        result = result + str(cur.get_element())  + ", "
        # proceed to next node
        cur = cur.get_next()
    #enclose the result in a parantheses
    result = "(" + result[:-2] + ")"
    #return the result
    return result

如您所见，已经有在头部添加和在尾部添加的功能，但是我不知道如何在列表中间添加。我需要创建一个获取新数据的函数，并在单链表中间添加一个带有数据的节点。有人可以向我展示代码或如何修改这些功能之一或添加新功能吗？感谢您的帮助！

Answer 1

您必须从第一个元素开始，然后遍历列表，直到到达要插入的位置。假设您需要将节点node_to_insert 添加到位置pos：

current_node = self.get_head()
for i in range(pos): # iterate pos time:
    current_node = current_node.get_next() # Jump to next node
# Now curr node is `pos`-th node. We insert `node_to_insert` here.  
# Suppose u already have it named `node_to_insert`.  
# We need to store the next node (of current node),  
# so we can link the inserted node to it
next_current_node = current_node.get_next()
current_node.set_next(node_to_insert)
node_to_insert.set_next(next_current_node)

就是这样。希望这有帮助。我没有测试代码，但我希望你明白我的想法。

Answer 2

这应该可行：

class SingleLinkedList():

    ...

    def add_after(self, n, element):
        """add element after the n-th node, counting from 1 (if n=0, add before the first node)"""
        # check if we should add at the start...
        if (n<=0):
            self.add_first(element)
        # ... or the end
        elif (n>=self.size()):
            self.add_last(element)
        # otherwise, add after node n
        else:
            # start at the first node
            node = self.get_head()
            # loop over nodes until we reach node n
            for i in range(n - 1):
                node = node.get_next()
            # make a new node from our element
            new_node = Node(element,node.get_next())
            # insert it after node n
            node.set_next(new_node)

测试：

# test
q = SingleLinkedList()
q.add_first('A')
q.add_last('B')
q.add_last('C')
q.add_last('D')
q.add_last('E')
q.add_after(2, 'X')

print(q) # (A, B, X, C, D, E)