用Python实现链表Linklist

在leetcode上面试着用Python解题，但是遇到链表、二叉树什么的，Python就用不溜。在网上看了一些资料。

完整的实现过程如下：

class Node:    next = None    data = None        def __init__(self, nodeData):        self.data = nodeData# 一个链表数据结构，包括一个 根节点root 还有 链表大小sizeclass LinkList:    root = None    size = 0;            def __init__(self):        self.root = None        size = 0            def __del__(self):        "删除链表（清空节点）"        if self.root is None:            return        curNode = self.root        while curNode.next is not None:            tempNode = curNode            curNode = curNode.next            tempNode = None        curNode = None      # Insert a new node to the LinkList    def Insert(self, newData):        "添加节点"        newNode = Node(newData)  # 每次都新建一个Node（新地址）        if self.root is None:            self.root = newNode            self.size = 1            return                tempNode = self.root        while tempNode.next is not None:            tempNode = tempNode.next        tempNode.next = newNode        self.size += 1            # def Get data of Position pos    def GetData(self, pos):        if pos >= self.size or pos < 0:            return None        else:            tempNode = self.root            for i in range(0, pos):                tempNode = tempNode.next            return tempNode.data        # Remove a certain node    def Remove(self, theData):        curNode = self.root        if curNode is None:            return                if self.size == 1 and curNode.data == theData:            curNode.data = None            curNode = None            self.size -= 1            return        elif curNode.data == theData:  # 第一个节点就是需要删除的节点            self.root = curNode.next            curNode = None            self.size -= 1            return                    while curNode.next is not None:            if curNode.next.data == theData:                tempNode = curNode.next                curNode.next = curNode.next.next                tempNode = None  # remove the node,but curNode stays still                self.size -= 1            else:                curNode = curNode.next                    # Get Root Node    def GetRoot(self):        return self.root            # def Get Size    def GetSize(self):        return self.size        # print the data of the LinkList    def Print(self):        tempNode = self.root        while tempNode is not None:            print tempNode.data,            tempNode = tempNode.next                      if __name__ == '__main__':        print ("start main()")    mylist = LinkList()  # 一个LinkList 就是一个完整的链表; Node 为链表的节点    print mylist.GetSize()        for i in range(1, 10):        mylist.Insert(i)            print ('输出该单链表：')    mylist.Print()        print ("\n mylist 的 size:%s" % mylist.GetSize())        print ("---------------------------------\n删除1号索引的元素：")    mylist.Remove(6)       mylist.Remove(1)          print ("输出删除1号索引后链表的元素：\n")    mylist.Print()         print ("\n删除后的元素个数：%s" % mylist.GetSize())

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