python实现一个无序单链表

Posted 2020-12-08 RTFM

class Node:
    """先定一个node的类"""

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

    def getValue(self):
        return self.value

    def getNext(self):
        return self.next

    def setValue(self, new_value):
        self.value = new_value

    def setNext(self, new_next):
        self.next = new_next


class LinkedList:
    """实现一个单向链表及其各类操作方法"""

    def __init__(self):
        """初始化空链表"""
        self._head = Node()
        self._tail = None
        self._length = 0

    def isEmpty(self):
        """检测是否为空"""
        return self._head is None

    def add(self, value):
        """add在链表前端添加元素:O(1)"""
        newnode = Node(value)
        newnode.setNext(self._head)
        # 注意这里的顺序不能和setNext()颠倒不然新增的节点找不到next
        self._head = newnode

    def append(self, value):
        """append在链表尾部添加元素:O(n)
        思路：遍历链表，在原链尾next指向新节点"""
        newnode = Node(value)
        if self.isEmpty():
            # 若为空表，将添加的元素设为第一个元素
            self._head = newnode
        else:
            # 从链首遍历链表
            current = self._head
            while current.getNext() is not None:
                current = current.getNext()
            # 找到最后一个，直接设置它的next指向新增的节点
            current.setNext(newnode)

    def size(self):
        """获取链表的元素个数
        从链头head开始遍历到链尾，同时用变量累加经过的节点个数"""
        current = self._head
        while current is not None:
            current = current.getNext()
            self._length += 1
        return self._length

    def search(self, value):
        """查找元素是否在链表，找到返回True，否则返回False
        从链头head开始遍历到链尾，并判断当前节点的数据是否为目标value"""
        current = self._head
        found = False
        while current is not None and not found:
            if current.getValue() == value:
                found = True
            else:
                current = current.getNext()
        return found

    def remove(self, value):
        """删除一个元素
        遍历链表"""
        current = self._head
        previous = None
        found = False

        while not found:
            if current.getValue() == value:
                found = True
                # 找到后判断value是不是链首，是的话，head为value的下个节点
                if not previous:
                    self._head = current.getNext()
                else:
                    # 不是的话，将前一个节点的next指向要删除的节点的下一个节点
                    previous.setNext(current.getNext())
            elif current.getNext() is not None:
                # 之前的节点指向当前节点
                previous = current
                # 并找下一个节点作为循环的当前节点
                current = current.getNext()
            else:
                raise ValueError(‘{} is not in LinkedList‘.format(value))

    def index(self, value):
        """返回元素在链表的位置，找不到抛出ValueError错误
        遍历链表，并用count累加遍历过的每一个节点位置"""
        current = self._head
        count = 0
        found = False

        while current is not None and not found:
            if current.getValue() == value:
                found = True
            else:
                current = current.getNext()
                count += 1
        if found:
            return count
        else:
            raise ValueError(‘{} is not in LinkedList‘.format(value))

    def insert(self, position, value):
        """往链表position位置插入一个元素value"""
        # 如果是链首，直接add添加
        if position <= 1:
            self.add(value)
        # 如果是链尾，直接append
        elif position > self.size():
            self.append(value)
        # 中间位置插入，思路也是从头遍历，找到position位置之前一个节点插入
        # 并修改previous节点next指向新节点，新节点next指向position位置的节点
        else:
            temp = Node(value)
            previous = None
            count = 1
            current = self._head
            while count < position:
                count += 1
                previous = current
                current = current.getNext()

            previous.setNext(temp)
            temp.setNext(current)


if __name__ == ‘__main__‘:
    link = LinkedList()
    link.add(4)
    link.add(5)
    link.add(6)
    link.add(7)
    print(link.remove(4))
    print(link.size())