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表 (python实现)

时间:2019-11-29 10:54:25      阅读:77      评论:0      收藏:0      [点我收藏+]

标签:turn   int   app   设置   创建   相等   移动   link   node   

python 中的顺序表

  在python中的顺序表其实就是list。是动态数组(Cpython)。

  

  从细节上看,Python中的列表是由对其它对象的引用组成的连续数组。指向这个数组的指针及其长度被保存在一个列表头结构中。这意味着,每次添加或删除一个元素时,由引用组成的数组需要该标大小(重新分配)。幸运的是,Python在创建这些数组时采用了指数分配,所以并不是每次操作都需要改变数组的大小。但是,也因为这个原因添加或取出元素的平摊复杂度较低。
  技术图片

单链表

 

class Node(object):
    def __init__(self, item):

        self.elem = item
        self.next = None  #下一个节点的内存地址


class singleLinkList():
    def __init__(self):
        self.__head = None    #私有属性 有限范围在此类中。 链表头只存下一个节点的内存地址

    def is_empty(self)-> bool:

        return self.__head is None  #is 判断两变量内存地址是否相等

    def length(self)-> int:
        if self.is_empty():
            return 0
        count = 0
        cur = self.__head

        while cur is not self.__head:   #while cur
            cur += 1
            cur = cur.next
        return count

    def add(self, item):
        #头插法
        node = Node(item)
        node.next = self.__head
        self.__head = node

    def travel(self):
        if self.is_empty():
            return
        cur = self.__head
        while cur:
            print(cur.elem)
            cur = cur.next

    def append(self,item):
        if self.is_empty():
            self.add(item)
        else:
            cur = self.__head
            node = Node(item)
            while cur:
                cur = cur.next
            cur=node
    def insert(self,index:int,item:int):
        if index <= 0:
            self.add(item)
        elif index>self.length()-1:
            self.append(item)
        else:
            cur = self.__head
            n=1
            while cur.next:
                if n==index:
                    break
                n += 1
                cur = cur.next
        node = Node(item)
        node.next = cur.next  # 被插节点next 赋给 node‘s next
        cur.next = node       # 被插节点的next 指向 node

    def remove(self,item):
        if self.is_empty:  # 节点为空的情况
            raise ValueError("null")
        cur = self.__head
        pre = None
        if self.item == item:
            self.__head = cur.next
        while cur.next:
            pre=cur
            cur=cur.next
            if cur.item == item:
                pre.next = cur.next

 

 循环单链表

class Node():

    def __init__(self, item):
        self.item = item
        self.next = None


# 单向循环链表
class SingleLink():

    def __init__(self):
        self.__head = None

    def is_empty(self):

        return self.__head is None

    def length(self):
        if self.is_empty():
            return 0
        n = 1
        cur = self.__head
        while cur.next != self.__head:
            cur = cur.next
            n += 1
        return n

    def travel(self):
        if self.is_empty():
            return False
        cur = self.__head
        while cur.next != self.__head:
            print(cur.item)
            cur = cur.next

    def add(self, item):
        """首插法
        """
        node = Node(item)
        if self.is_empty():
            self.__head = node
            node.next = self.__head
        else:
            cur = self.__head
            node.next = self.__head  # 插入节点先指向head
            while cur.next != self.__head:
                cur = cur.next
            cur.next = node
            self.__head = node

    def append(self, item):
        """尾插法
        """
        node = Node(item)
        if self.is_empty():
            self.__head = node
            node.next = self._head
        else:
            cur = self.__head
            while cur.next != self.__head:
                cur = cur.next
            cur.next = node
            node.next = self.__head

    def insert(self, pos, item):
        """
        插入节点
        :param  pos item:
        :return:
        """
        if pos < 0:
            self.add(item)
        elif pos > (self.length()-1):
            self.append(item)
        else:
            node = Node(item)
            cur = self.__head
            n = 1
            while n != pos:
                cur = cur.next
                n += 1
            node.next = cur.next
            cur.next = node

    def remove(self, item):
        """
        删除节点
        :param item:
        :return:
        """
        if self.is_empty():
            return
        cur = self.__head
        pre = None
        if cur.item == item:
            # 如果链表不止一个节点
            if cur.next != self._head:
                # 先找到尾节点,将尾节点的next指向第二个节点
                while cur.next != self._head:
                    cur = cur.next
                # cur 指向了尾节点
                cur.next = self._head.next
                self._head = self._head.next
            else:
                # 链表只有一个节点
                self._head = None
            # 头节点不是要删除的元素
        else:
            pre = self._head
            while cur.next != self._head:
                # 找到了要删除的元素
                if cur.item == item:
                    pre.next = cur.next
                    return
                else:
                    pre = cur
                    cur = cur.next
            # 尾节点操作
            if cur.item == item:
                #尾部删除
                pre.next = cur.next

    def search(self, item):
        """查找节点是否存在"""
        if self.is_empty():
            return False
        cur = self.__head
        if cur.item == item:
            return True
        while cur.next != self.__head:
            cur = cur.next
            if cur.item == item:
                return True
        return False

 

 双向链表

class Node():

    def __init__(self, item):
        self.item = item
        self.prev =None
        self.next = 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 is None

    def length(self):
        """返回链表的长度"""
        cur = self.__head
        count = 0
        while cur != None:
            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):
        """头插法"""
        node = Node(item)
        if self.is_empty():
            self.__head = node
        else:
            node.next = self.__head
            self.__head.prev = 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
            # 操作流程为顺时针1234
            # 1.将node的prev指向cur
            node.prev = cur
            # 2.将node的next指向cur的下一个节点
            node.next = cur.next
            # 3.将cur的下一个节点的prev指向node
            cur.next.prev = node
            # 4.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
            # 如果首节点的元素就是要删除的元素
            if cur.item == item:
                # 如果链表只有这一个节点
                if cur.next == None:
                    self._head = None
                else:
                    # 将第二个个节点的prev设置为None
                    cur.next.prev = None
                    # 将_head指向第二个节点
                    self._head = cur.next
                return
            # 首节点不是要删除的节点
            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
            # 尾结点删除

 

 

表 (python实现)

标签:turn   int   app   设置   创建   相等   移动   link   node   

原文地址:https://www.cnblogs.com/sometingintheway/p/11915233.html

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