算法-02-双链表

# 双向链表,每个结点都有两个指针域
class Node(object):
    def __init__(self, item):
        self.elem = item
        self.next = None # 定义为指针
        self.prev = None # 定义头指针

class DoubleLinkList(object):
    def __init__(self, node=None):
        self.__head = node

    def is_empty(self):
        return self.__head == None

    def length(self):
        count = 0
        cur = self.__head
        while cur != None:
            count += 1
            cur = cur.next
        return count
    def travel(self):
        cur = self.__head
        while cur != None:
            print(cur.elem)
            cur = cur.next
    def add(self, item):
        node = Node(item)
        node.next = self.__head
        self.__head.prev = node
        self.__head = node
    def append(self, item):
        node = Node(item)
        if self.is_empty():
            self.__head = node
        else:
            cur = self.__head
            while cur is not None:
                cur.next = node
                node.prev = cur
    def insert(self, pos, item):
        node = Node(item)
        if pos <= 0:
            self.add(item)
        elif pos >= self.length()-1:
            self.append(item)
        else:
            cur = self.__head
            count = 0
            while count != pos:
                count += 1
                cur = cur.next
            '''方法一:
            node.next = cur # 使新结点的尾指针指向插入位置的下一个结点
            node.prev = cur.prev # 使新结点的头指针指向插入位置的前一个结点,用这两步实现新结点与前后结点之间的链接,为后面拆开原有前后结点之间的关系做铺垫
            cur.prev.next = node # 使插入位置的前结点的尾指针指向新结点,从而拆开原有前后结点之间的链接
            cur.prev = node # 使插入位置的头指针指向新结点,实现新结点与原前后结点最后的链接'''
            #方法二:
            node.next = cur
            node.prev = cur.prev
            cur.prev = node
            cur.prev.next = node
            # 二者连接顺序都是:先使新结点node的头指针与尾指针,与想插入位置的前后结点构成连接,当连接好后再思考如何拆开原有两个结点之间的连接
    def remove(self, item):
        cur = self.__head
        while cur != None:
            if cur.elem == item:
                if cur == self.__head:
                    self.__head = cur.next
                    # 判断链表是否只有一个结点
                    if cur.next:
                        cur.next.prev = None
                break
            else:
                cur.prev.next = cur.next
                # 判断是否是在尾结点进行的删除,当cur.next存在时返回为真
                if cur.next:
                    cur.next.prev = cur.prev
                break
    def search(self, item):
        cur = self.__head
        while cur != None:
            if cur.elem == item:
                return True
            else:
                cur = cur.next
            return False