python中的单向循环链表实现

绝地灬酷狼 2021-09-20 11:32 284阅读 0赞

# 引子 #

![1190058-20180428094824956-14592460.png][]

所谓单向循环链表，不过是在单向链表的基础上，如响尾蛇般将其首尾相连，也因此有诸多类似之处与务必留心之点。尤其是可能涉及到头尾节点的操作，不可疏忽。

对于诸多操所必须的遍历，这时的条件是什么？又应该在哪里停止？

在做删除操作时，如若待删除节点是头或尾节点时，该如何处理？如果链表只有一个节点，又该如何处理？

# 代码实现 #

class Node(object):
        def __init__(self, value):
            # 元素域
            self.value = value
            # 链接域
            self.next = None
    
    
    class CircularLinkedListOneway(object):
        def __init__(self, node=None):
            # 构造非空链时，让其地址域指向自己
            if node:
                node.next = node
            self.__head = node
    
        def is_empty(self):
            # 头节点不为None则不为空
            return self.__head == None
    
        def __len__(self):
            count = 1
            cur = self.__head
            if self.is_empty():
                return 0
            while cur.next != self.__head:
                count += 1
                cur = cur.next
            return count
    
        def traversal(self):
            if self.is_empty():
                return
            cur = self.__head
            while cur.next != self.__head:
                print(cur.value)
                cur = cur.next
            # 退出循环时，cur正是尾节点
            print(cur.value)
    
        def add(self, value):
            """头插法"""
            node = Node(value)
            if self.is_empty():
                self.__head = node
                self.__head.next = self.__head
                return
            cur = self.__head
            while cur.next != self.__head:
                cur = cur.next
            # 新节点的next指针指向原头节点
            node.next = self.__head
            # 将新节点指向头节点
            self.__head = node
            # 尾节点next指针指向新头节点
            cur.next = self.__head
    
        def append(self, value):
            node = Node(value)
            cur = self.__head
            if self.is_empty():
                self.__head = node
                self.__head.next = self.__head
                return
            while cur.next != self.__head:
                cur = cur.next
            node.next = cur.next
            cur.next = node
    
        def insert(self, pos, value):
            if pos <= 0:
                self.add(value)
            elif pos > len(self) - 1:
                self.append(value)
            else:
                node = Node(value)
                cur = self.__head
                count = 0
                while count < pos - 1:
                    count += 1
                    cur = cur.next
                node.next = cur.next
                cur.next = node
    
        def search(self, value):
            if self.is_empty():
                return False
            cur = self.__head
            while cur.next != self.__head:
                if cur.value == value:
                    return True
                else:
                    cur = cur.next
            # 别忘了while循环外的尾节点
            if cur.value == value:
                return True
            return False
    
        def remove(self, value):
            cur = self.__head
            prior = None
            if self.is_empty():
                return
            while cur.next != self.__head:
                # 待删除节点如果找到
                if cur.value == value:
                    # 待删除节点在头部
                    if cur == self.__head:
                        rear = self.__head
                        while rear.next != self.__head:
                            rear = rear.next
                        self.__head = cur.next
                        rear.next = self.__head
                    # 待删除节点在中间
                    else:
                        prior.next = cur.next
    　　　　　　　　　 # 这里不是跳出循环的break,而是退出函数的return哦,因为已经处理完毕了
                    return
                # 如果还没找到
                else:
                    prior = cur
                    cur = cur.next
            # 待删除节点在尾部
            if cur.value == value:
                # 如果链表中只有一个元素，则此时prior为None，Next属性就会报错
                # 此时直接使其头部元素为None即可
                if cur == self.__head:
                    self.__head = None
                    return
                prior.next = cur.next

转载于:https://www.cnblogs.com/yifeixu/p/8965082.html

[1190058-20180428094824956-14592460.png]: /images/20210726/8214084ee8e846e9affef18f7b4bb64b.png