编程小白求指导,如何实现如图所示的数据结构,并且统计节点数
1条回答
- 无明之徒 2018-04-08 10:45关注
python实现链表
首先,定义节点类Node:
class Node:
'''
data: 节点保存的数据
next: 保存下一个节点对象
'''
def __init_(self, data, pnext=None):
self.data = data
self._next = pnextdef __repr__(self): ''' 用来定义Node的字符输出, print为输出data ''' return str(self.data) 然后,定义链表类:
链表要包括:
属性:
链表头:head
链表长度:length
方法:
判断是否为空: isEmpty()
def isEmpty(self):
return (self.length == 0
增加一个节点(在链表尾添加): append()
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 删除一个节点: delete() #删除一个节点之后记得要把链表长度减一
def delete(self, index):
if self.isEmpty():
print "this chain table is empty."
returnif index < 0 or index >= self.length: print 'error: out of index' return #要注意删除第一个节点的情况 #如果有空的头节点就不用这样 #但是我不喜欢弄头节点 if index == 0: self.head = self.head._next self.length -= 1 return #prev为保存前导节点 #node为保存当前节点 #当j与index相等时就 #相当于找到要删除的节点 j = 0 node = self.head prev = self.head while node._next and j < index: prev = node node = node._next j += 1 if j == index: prev._next = node._next self.length -= 1 修改一个节点: update() def update(self, index, data): if self.isEmpty() or index < 0 or index >= self.length: print 'error: out of index' return j = 0 node = self.head while node._next and j < index: node = node._next j += 1 if j == index: node.data = data 查找一个节点: getItem() def getItem(self, index): if self.isEmpty() or index < 0 or index >= self.length: print "error: out of index" return j = 0 node = self.head while node._next and j < index: node = node._next j += 1 return node.data 查找一个节点的索引: getIndex() def getIndex(self, data): j = 0 if self.isEmpty(): print "this chain table is empty" return node = self.head while node: if node.data == data: return j node = node._next j += 1 if j == self.length: print "%s not found" % str(data) return 插入一个节点: insert() 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 j = 0 node = self.head prev = self.head while node._next and j < index: prev = node node = node._next j += 1 if j == index: item._next = node prev._next = item self.length += 1 清空链表: clear() def clear(self): self.head = None self.length = 0
解决 无用评论 打赏 举报
悬赏问题
- ¥15 如何用Labview在myRIO上做LCD显示?(语言-开发语言)
- ¥15 Vue3地图和异步函数使用
- ¥15 C++ yoloV5改写遇到的问题
- ¥20 win11修改中文用户名路径
- ¥15 win2012磁盘空间不足,c盘正常,d盘无法写入
- ¥15 用土力学知识进行土坡稳定性分析与挡土墙设计
- ¥70 PlayWright在Java上连接CDP关联本地Chrome启动失败,貌似是Windows端口转发问题
- ¥15 帮我写一个c++工程
- ¥30 Eclipse官网打不开,官网首页进不去,显示无法访问此页面,求解决方法
- ¥15 关于smbclient 库的使用