python单链表

内容导读

互联网集市收集整理的这篇技术教程文章主要介绍了python单链表，小编现在分享给大家，供广大互联网技能从业者学习和参考。文章包含2550字，纯文字阅读大概需要4分钟。

内容图文

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

    def __repr__(self):
        return str(self.data)

class Chain:
    def __init__(self):
        self.head = None
        self.length = 0

    def isEmpty(self):
        return self.length == 0

    def append(self, dataOrNode):
        item = None
        if isinstance(dataOrNode, Node):
            item = dataOrNode
        else:
            item = Node(dataOrNode)

        if not self.head:
            self.head = item
            self.length += 1

        else:
            node = self.head
            while node._next:
                node = node._next
            node._next = item
            self.length += 1

    def delete(self, index):
        if self.isEmpty():
            print("this chain table is empty.")
            return

        if index < 0 or index >= self.length:
            print('error: out of index')
            return

        if index == 0:
            self.head = self.head._next
            self.length -= 1
            return

        node = self.head
        prev = self.head
        for i in range(index):
            prev = node
            node = node._next
        prev._next = node._next
        self.length -= 1

    def insert(self, index, dataOrNode):
        if self.isEmpty():
            print("this chain tabale is empty")
            return

        if index < 0 or index >= self.length:
            print("error: out of index")
            return

        item = None
        if isinstance(dataOrNode, Node):
            item = dataOrNode
        else:
            item = Node(dataOrNode)

        if index == 0:
            item._next = self.head
            self.head = item
            self.length += 1
            return

        node = self.head
        prev = self.head
        for i in range(index):
            prev = node
            node = node._next
        item._next = node
        prev._next = item
        self.length += 1

    def update(self, index, data):
        if self.isEmpty() or index < 0 or index >= self.length:
            print('error: out of index')
            return

        node = self.head
        for i in range(index):
            node = node._next
        node.data = data

    def getItem(self, index):
        if self.isEmpty() or index < 0 or index >= self.length:
            print("error: out of index")
            return
        node = self.head
        for i in range(index):
            node = node._next
        return node.data

    def getIndex(self, data):
        i = 0
        if self.isEmpty():
            print("this chain table is empty")
            return
        node = self.head
        while node:
            if node.data == data:
                return i
            node = node._next
            i += 1

        if i == self.length:
            print("%s not found" % str(data))
            return

    def clear(self):
        self.head = None
        self.length = 0

    def __repr__(self):
        if self.isEmpty():
            return "empty chain table"
        node = self.head
        nlist = ''
        while node:
            nlist += str(node.data) + ' '
            node = node._next
        return nlist

    def __getitem__(self, ind):
        if self.isEmpty() or ind < 0 or ind >= self.length:
            print("error: out of index")
            return
        return self.getItem(ind)

    def __setitem__(self, ind, val):
        if self.isEmpty() or ind < 0 or ind >= self.length:
            print("error: out of index")
            return
        self.update(ind, val)

    def __len__(self):
        return self.length



if __name__ == "__main__":
    c = Chain()
    for i in range(10):
        c.append(i)
    print(c)

• 点赞
• 收藏
• 分享
• • 文章举报

内容总结

以上是互联网集市为您收集整理的python单链表全部内容，希望文章能够帮你解决python单链表所遇到的程序开发问题。 如果觉得互联网集市技术教程内容还不错，欢迎将互联网集市网站推荐给程序员好友。

内容备注

版权声明：本文内容由互联网用户自发贡献，该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容， 请发送邮件至 gblab@vip.qq.com 举报，一经查实，本站将立刻删除。

内容手机端