Question

我的代码如下。它在循环列表中附加了一些数字。我的程序运行正常。它给出了[5,3,3]或输入的任何数字的精确输出。但我想在输出中做一些改变。在没有添加任何新函数的情况下，在def append(....)和def add_before(...)中进行了哪些更改，因此它提供了一个唯一的数字，这意味着它可以删除重复项。例如，将给出[5,3]

class CirList:
    def __init__(self):
        head_node = NodeDLL(None)
        head_node.next = head_node
        head_node.prev = head_node
        self.__head = head_node

    def append(self, item):
        curr = self.__head
        new_node = NodeDLL(item, curr, curr.get_prev())
        curr.set_prev(new_node)
        new_node.get_prev().set_next(new_node)

    def add_before(self, item, old_item):
        curr = self.__head.next
        found = False
        while curr.get_data() != None and not found:
            if curr.get_data() == old_item:
                found = True  
            else:
                curr = curr.get_next()
        if found:
            new_node = NodeDLL(item, curr, curr.get_prev())
            curr.set_prev(new_node)
            new_node.get_prev().set_next(new_node)
        return found
    def remove(self, item):
        curr = self.__head.next
        found = False
        while curr.get_data() != None and not found:
            if curr.get_data() == item:
                found = True
            else:
                curr = curr.get_next()
        if found:       
            curr.get_prev().set_next(curr.get_next())
            curr.get_next().set_prev(curr.get_prev())
    def printall(self):
        curr = self.__head.next
        while curr.get_data() != None:
            print(curr.get_data(), end=" ")
            curr = curr.get_next()
        print()
    def __str__(self):
        result = "["
        curr = self.__head.next 
        while curr.get_data() != None:
            result += str(curr.get_data()) + " "
            curr = curr.get_next()
        result = result.rstrip(" ")
        result += "]"
        return result

测试

listA = CirList()
listA.append(5)
listA.append(3)
listA.append(3)
print(listA)

Answer 1

两个选项（我能想到）：

不要添加重复项：

class CirList:
    def __init__(self):
        head_node = NodeDLL(None)
        head_node.next = head_node
        head_node.prev = head_node
        self.__head = head_node
        self._knownNumbers = set() # optimized lookup if number known

    def append(self, item):
        if item not in self._knownNumbers: # only add if not known
            self.__knownNumbers__.add(item)
            curr = self.__head
            new_node = NodeDLL(item, curr, curr.get_prev())
            curr.set_prev(new_node)
            new_node.get_prev().set_next(new_node)

    def add_before(self, item, old_item):
        if item not in self._knownNumbers: # only add if not known
            self.__knownNumbers__.add(item)
            curr = self.__head.next
            found = False
            while curr.get_data() != None and not found:
                if curr.get_data() == old_item:
                    found = True  
                else:
                    curr = curr.get_next()
            if found:
                new_node = NodeDLL(item, curr, curr.get_prev())
                curr.set_prev(new_node)
                new_node.get_prev().set_next(new_node)
            return found
    def remove(self, item):
        self._knownNumbers.remove(item) # forget this number again
        curr = self.__head.next
        found = False
        while curr.get_data() != None and not found:
            if curr.get_data() == item:
                found = True
            else:
                curr = curr.get_next()
        if found:       
            curr.get_prev().set_next(curr.get_next())
            curr.get_next().set_prev(curr.get_prev())

或者只是不打印重复：

def __str__(self):
    result = "["
    curr = self.__head.next
    known = set() # keep what we added already
    while curr.get_data() != None:
        if curr.get_data() not in known: # only add if not yet added
            result += str(curr.get_data()) + " "
            known.add(curr.get_data())   # remember this one
        curr = curr.get_next()
    result = result.rstrip(" ")
    result += "]"
    return result

如果你想让它模仿这种行为你必须相应地修改你的printall() - 你仍然会存储所有重复项，但这对我来说没有多大意义，除非你创建一个单独的def printNoDuplicates(self)特别为此目的。

如何从循环链表中删除重复项

1 个答案: