在python中删除链表中的节点 [英] Deleting a node in Linked List in python

问题描述

删除链表中的一个节点，这个实现有什么问题?:

To delete a node in a linked list, what is wrong with this implementation?:

def delete(self, val):
    tmp = self.head
    prev = None
    while tmp: 
        if val == tmp.data:
            self.size -= 1
            if prev==None: 
                self.head = self.head.next 
            else:
                prev.next = tmp.next 
        else:
            prev = tmp
            tmp = tmp.next 

我看过的所有指南都说应该是:

All the guides I've looked at say it should be:

def delete(self, data):
    tmp = self.head
    prev = None
    found = False 
    while tmp and not found:
        if data == tmp.data:
            found = True
        else:
            prev = tmp
            tmp = tmp.next
    if found:
        self.size -= 1
        if prev == None:
            self.head = self.head.next
        else:
            prev.next = tmp.next

但我不明白为什么需要found.为什么发现是必要的?为什么这个实现更正确?

but I can't figure out why the found is needed. Why is the found necessary? Why is this implementation more correct?

另外，我在搜索时也遇到了同样的问题:

Additionally, I have the same trouble with search:

我的实现是:

def __contains__(self, data):
    tmp = self.head
    while tmp:
        if data == tmp.data:
            return True 
        else: 
            tmp = tmp.next
    return False 

但正确的实现是:

def __contains__(self, data):
        tmp = self.head
        found = False
        while tmp and not found:
            if data == tmp.data:
                found = True
            else: 
                tmp = tmp.next
        return found 

推荐答案

只要数据是唯一的，delete 就是相同的.因此，通常更好的是，从处理元素中对列表进行单独的循环.它更具可读性，更少嵌套.如果找不到 data ，最好给出一个错误:

The deletes are identical as long as data is unique. So it is in general better, do separate looping through the list, from working on elements. It is more readable, and less nested. It would be still better, to give an error, if data is not found:

def delete(self, data):
    tmp = self.head
    prev = None
    while tmp:
        if data == tmp.data:
            break
        prev = tmp
        tmp = tmp.next
    else:
        raise ValueError('data not found')
    self.size -= 1
    if prev is None:
        self.head = tmp.next
    else:
        prev.next = tmp.next

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