Linus: 这样写是不懂指针

329次阅读  |  发布于2年以前

但凡刷过leetcode或者参加面试,大抵都遇到过如下这种问题:删除单链表中value为给定值的所有节点

假设链表节点定义如下:

struct ListNode {
  int value;
  ListNode* next;
};

那么,为了实现该功能,我们需要从头到尾遍历链表,同时删除value等于给定值的节点,显然,我们会像如下这样写:

void Remove(ListNode *head, int to_remove) {
  ListNode *entry = head;
  ListNode *prev = NULL;

  while (entry) {
    if (entry->val == to_remove) {
        if (prev)
           prev->next = entry->next;
        else
           head = entry->next;} else {
    prev = entry; }
    entry = entry->next;
  }
}

代码实现中,我们所做的是遍历列表,直到entry为 NULL(第5行)。当我们遇到要删除的节点时(第 6 行),我们将当前next指针的值分配给前一个,进而从当前链表中摘除当前元素(第 8 行)。

上述这种写法,相对来说简单、易理解,但 Linus Torvalds却不这么认为,其曾经评价这种写法:

At the opposite end of the spectrum, I actually wish more people understood the really core low-level kind of coding. Not big, complex stuff like the lockless name lookup, but simply good use of pointers-to-pointers etc. For example, I’ve seen too many people who delete a singly-linked list entry by keeping track of the “prev” entry, and then to delete the entry, doing something like

if (prev) prev->next = entry->next; else list_head = entry->next;

and whenever I see code like that, I just go “This person doesn’t understand pointers”. And it’s sadly quite common.

People who understand pointers just use a “pointer to the entry pointer”, and initialize that with the address of the list_head. And then as they traverse the list, they can remove the entry without using any conditionals, by just doing a “*pp = entry->next”

其大概意思就是,对于写出本文一开始那种代码的评价就是:***这个人不懂指针(This person doesn’t understand pointers)***。

那么,Linus推荐的写法又是什么样的呢?

void Remove(ListNode *head, int to_remove) {
  ListNode **pp = &head;
  ListNode *entry = head;

  while (entry) {
    if (entry->val == to_remove)
        *pp = entry->next;
    else
          pp = &entry->next;
    entry = entry->next;
  }
}

这块代码看起来相对简洁,但是,如果第一次遇到这种实现,可能需要花一点时间进行理解~~

实际上刷过题的都知道,list相关问题首先在栈上声明一个dummy node,能简化各种插入删除操作代码。使用dummy node操作后,代码如下:

void Remove(ListNode *head, int to_remove) {
  ListNode tmp;
  tmp.next = head;
  ListNode *pre = tmp;
  while (pre->next) {
    if (pre->next->val == to_remove) {
      pre->next = pre->next->next;
    } else {
      pre = pre->next;
    }
  }
}

相比来说,较Linus的实现方式更为简洁,但明显的,需要生成一个dummy节点,最明显的,内存使用上并不是最优,尽管其更易于理解和实现。

链表属于LC中相对来说比较简单的部分,选择一种适合自己的方式,并不一定非要选择最优,选择一个适合自己的。至于Linus实现的这种,可以保留意见,且当其最优吧,毕竟:没有人比Linus更懂指针,哈哈哈~

好了,今天的文章就到这,我们下期见!

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