用Python实现链表Linklist

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size
class 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())

参考：http://blog.csdn.net/anialy/article/details/7699509