Python实现双链表

双向链表（Double_linked_list）也叫双链表，是链表的一种，它的每个数据结点中都有两个指针，分别指向直接后继和直接前驱。所以，从双向链表中的任意一个结点开始，都可以很方便地访问它的前驱结点和后继结点。

这里直接给出参考代码，大家有兴趣请自行探索，这里就不再详细介绍。

在/home/shiyanlou/下新建一个文件double_linked_list.py。

具体实现代码如下：

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class Node(object):
    # 双向链表节点
    def __init__(self, item):
        self.item = item
        self.next = None
        self.prev = None
class DLinkList(object):
    # 双向链表
    def __init__(self):
        self._head = None
    def is_empty(self):
        # 判断链表是否为空
        return self._head == None
    def get_length(self):
        # 返回链表的长度
        cur = self._head
        count = 0
        while cur != None:
            count=count+1
            cur = cur.next
        return count
    def travel(self):
        # 遍历链表
        cur = self._head
        while cur != None:
            print(cur.item)
            cur = cur.next
        print("")
    def add(self, item):
        # 头部插入元素
        node = Node(item)
        if self.is_empty():
            # 如果是空链表，将_head指向node
            self._head = node
        else:
            # 将node的next指向_head的头节点
            node.next = self._head
            # 将_head的头节点的prev指向node
            self._head.prev = node
            # 将_head 指向node
            self._head = node
    def append(self, item):
        # 尾部插入元素
        node = Node(item)
        if self.is_empty():
            # 如果是空链表，将_head指向node
            self._head = node
        else:
            # 移动到链表尾部
            cur = self._head
            while cur.next != None:
                cur = cur.next
            # 将尾节点cur的next指向node
            cur.next = node
            # 将node的prev指向cur
            node.prev = cur
    def search(self, item):
        # 查找元素是否存在
        cur = self._head
        while cur != None:
            if cur.item == item:
                return True
            cur = cur.next
        return False
    def insert(self, pos, item):
        # 在指定位置添加节点
        if pos <= 0:
            self.add(item)
        elif pos > (self.length()-1):
            self.append(item)
        else:
            node = Node(item)
            cur = self._head
            count = 0
            # 移动到指定位置的前一个位置
            while count < (pos-1):
                count += 1
                cur = cur.next
            # 将node的prev指向cur
            node.prev = cur
            # 将node的next指向cur的下一个节点
            node.next = cur.next
            # 将cur的下一个节点的prev指向node
            cur.next.prev = node
            # 将cur的next指向node
            cur.next = node
    def remove(self, item):
        # 删除元素
        if self.is_empty():
            return
        else:
            cur = self._head
            if cur.item == item:
                # 如果首节点的元素即是要删除的元素
                if cur.next == None:
                    # 如果链表只有这一个节点
                    self._head = None
                else:
                    # 将第二个节点的prev设置为None
                    cur.next.prev = None
                    # 将_head指向第二个节点
                    self._head = cur.next
                return
            while cur != None:
                if cur.item == item:
                    # 将cur的前一个节点的next指向cur的后一个节点
                    cur.prev.next = cur.next
                    # 将cur的后一个节点的prev指向cur的前一个节点
                    cur.next.prev = cur.prev
                    break
                cur = cur.next