Python实现单向链表 单向循环链表 双向链表

单向链表

class Node:
    def __init__(self,val):
        self.val = val
        self.next = None


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

    # 判断链表是否为空
    def is_empty(self):
        return self.__head == None

    # 获取链表长度
    def length(self):
        l = 0
        cur = self.__head
        while cur:
            l += 1
            cur = cur.next
        return l

    # 遍历链表
    def travel(self):
        cur = self.__head
        while cur:
            print(f'{cur.val}->',end='')
            cur = cur.next
        print('None')

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

    # 尾插法
    def append(self,item):
        cur = self.__head
        node = Node(item)
        # 判断当前链表是否为空,为空就直接让head指向node节点
        if self.is_empty():
            self.__head = node

        while cur.next != None:
            cur = cur.next
        cur.next = node

    # 指定位置添加元素
    def insert(self,index,item):
        # 如果index>=链表长度,就使用尾插发
        if index >= self.length():
            self.append(item)
            return
        # index<=0就使用头插法
        if index <= 0:
            self.add(item)
            return
        i = 1
        cur = self.__head
        node = Node(item)
        while i < index:
            cur = cur.next
            i += 1
        # 跳出循环之后,cur指向的就是index位置的前一个位置
        # 现将node的next指向cur的next
        # 再将cur的next指向node
        node.next = cur.next
        cur.next = node

    # 移除节点
    def remove(self,item):
        cur = self.__head
        pre = None
        while cur:
            # 判断当前节点的val是否等于item
            if cur.val == item:
                # 判断当前节点是否是头结点
                if cur == self.__head:
                    self.__head = cur.next
                else:
                    pre.next = cur.next
                return
            else:
                pre = cur
                cur = cur.next


    # 查询元素
    def search(self,item):
        cur = self.__head
        while cur:
            if cur.val == item:
                return True
            cur = cur.next
        return False

单向循环链表

class Node:
    def __init__(self,val):
        self.val = val
        self.next = None


class Sincyclinklist:
    def __init__(self):
        self.__head = None

    # 判断链表是否为空
    def is_empty(self):
        return self.__head == None

    # 获取链表长度
    def length(self):
        cur = self.__head
        if self.is_empty():
            return 0
        else:
            l = 1
            while cur.next != self.__head:
                l += 1
                cur = cur.next
            return l

    # 遍历链表
    def travel(self):
        cur = self.__head
        if self.is_empty():
            print('None')
        else:
            while cur.next != self.__head:
                print(f'{cur.val}->', end='')
                cur = cur.next
            print(cur.val)

    # 头插法
    def add(self, item):
        node = Node(item)
        if self.is_empty():
            self.__head = node
            node.next = node
        else:
            cur = self.__head
            while cur.next != self.__head:
                cur = cur.next
            # 跳出循环,cur指向的就是最后一个节点
            node.next = self.__head
            self.__head = node
            cur.next = node

    # 尾插法
    def append(self, item):
        cur = self.__head
        node = Node(item)
        # 判断当前链表是否为空,为空就直接让head指向node节点
        # 在让node.next指向自己或者self.__head都是可以的
        if self.is_empty():
            self.__head = node
            node.next = node
        else:
            while cur.next != self.__head:
                cur = cur.next
            # 跳出循环,cur指向的就是最后一个节点

            cur.next = node
            node.next = self.__head

    # 指定位置添加元素
    def insert(self, index, item):
        # 如果index>=链表长度,就使用尾插法
        if index >= self.length():
            self.append(item)
            return
        # index<=0就使用头插法
        if index <= 0:
            self.add(item)
            return
        i = 1
        cur = self.__head
        node = Node(item)
        while i < index:
            cur = cur.next
            i += 1
        # 跳出循环之后,cur指向的就是index位置的前一个位置
        # 先将node.next指向cur.next
        # 再将cur.next指向node
        node.next = cur.next
        cur.next = node

    # 删除节点
    def remove(self, item):
        if self.is_empty():
            return
        cur = self.__head
        pre = None
        while cur.next != self.__head:
            # 判断当前节点的val是否等于item
            if cur.val == item:
                # 判断当前节点是否是头结点
                if cur == self.__head:
                    # 创建一个新的游标,让它指向最后一个节点
                    new_cur = self.__head
                    while new_cur.next != self.__head:
                        new_cur = new_cur.next
                    # 最后一个节点再指向cur.next
                    # 让self.__head等于cur.next,也就是指向cur.next
                    new_cur.next = cur.next
                    self.__head = cur.next
                else:
                    pre.next = cur.next
                return
            # 如果当前节点不是要删除的元素,pre和cur就同时往后移动
            else:
                pre = cur
                cur = cur.next

        # 跳出循环之后,还有最后一个节点没有做判断
        # 删除最后一个节点
        if cur.val == item:
            # 判断当前链表是否只有一个节点
            # 只有一个节点的时候是进入不了上面的循环,所以说当前的pre还是为None,None是没有next这个属性的
            # 所以下面的判断依据也可以改成if not pre:
            if self.length() == 1:
                self.__head == None
            else:
                pre.next = self.__head


    # 查询元素
    def serach(self, item):
        # 判断是否为空
        if self.is_empty():
            return False
        cur = self.__head
        while cur.next != self.__head:
            if cur.val == item:
                return True
            cur = cur.next
        if cur.val == item:
            return True
        return False

双向链表

class Node:
    def __init__(self,val):
        self.val = val
        self.next = None
        self.pre = None

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

    # 判断链表是否为空
    def is_empty(self):
        return self.__head == None

    # 获取链表长度
    def length(self):
        l = 0
        cur = self.__head
        while cur:
            l += 1
            cur = cur.next
        return l

    # 遍历链表
    def travel(self):
        cur = self.__head
        while cur:
            # 只是为了让打印结果看起来像一个链表,你怎么打印都行
            print(f'{cur.val}<->', end='')
            cur = cur.next
        print('None')

    # 头插法
    def add(self,item):
        node = Node(item)
        if self.is_empty():
            self.__head = node
        else:
            node.next = self.__head
            # 把第一个节点的pre指向node
            self.__head.pre = node
            # 再将self.__head指向node
            self.__head = node


    # 尾插法
    def append(self,item):
        node = Node(item)
        if self.is_empty():
            self.__head == node
        else:
            cur = self.__head
            while cur.next != None:
                cur = cur.next
            cur.next = node
            node.pre = cur

    # 按位置插入
    def insert(self,index,item):
        # 如果index>=链表长度,就使用尾插发
        if index >= self.length():
            self.append(item)
            return
        # index<=0就使用头插法
        if index <= 0:
            self.add(item)
            return
        i = 0
        cur = self.__head
        node = Node(item)
        while i < index:
            cur = cur.next
            i += 1
        # 跳出循环之后,cur指向的就是index位置的节点
        # 把新节点的pre指向cur.pre
        node.pre = cur.pre
        # 把新节点的next指向cur
        node.next = cur
        # 再将cur.pre.next指向node
        cur.pre.next = node
        # 最后将node.pre指向node
        node.pre = node


    def remove(self,item):
        cur = self.__head
        while cur:
            # 判断当前节点的值是否等于item
            if cur.val == item:
                # 判断当前节点是否是第一个节点
                if cur == self.__head:
                    self.__head = cur.next
                    cur.next.pre = None
                else:
                    cur.pre.next = cur.next
                    # 加一个判断,是因为要删除的节点是最后一个时,就不用执行下面这一步了
                    # 也就是直接让prev.next指向cur.next,这时的cur.next就是None
                    if cur.next:
                        cur.next.pre = cur.pre
                return
            else:
                cur = cur.next

    def serach(self,item):
        cur = self.__head
        while cur:
            if cur.val == item:
                return True
            cur = cur.next
        return False
    原文作者:做个不一样的小丑
    原文地址: https://blog.csdn.net/hmh4640219/article/details/115494135
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
点赞