c - 此代码在 Xcode 上可以运行，但在 leetcode 编译器中失败

Question

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* removeElements(struct ListNode* head, int val) {

    struct ListNode *beforeNode = head;

    while(beforeNode != NULL){

        if(head == beforeNode && head->val == val){
            struct ListNode* q = head;
            head = q->next;
            beforeNode = head;
            free(q);
        }
        else if(beforeNode->next != NULL && beforeNode->next->val == val){
            struct ListNode *p = beforeNode->next;
            beforeNode = p->next;
            free(p);
        }

       else
           beforeNode = beforeNode->next;

    }

    return head;
}

Answer 1

This code works on Xcode but fails in leetcode compiler

好吧，该代码在任何平台上都无法运行。

您在 Xcode 上完成的测试一定是不完整的，因为代码在任何平台上都有未定义的行为。

看这个简单的例子:

#include <stdio.h>
#include <stdlib.h>

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

void printList(struct ListNode* p)
{
  if (p)
  {
    printf("%p %d", (void*)p, p->val);
    while(p->next)
    {
      printf("->");
      p = p->next;
      printf("%p %d", (void*)p, p->val);
    }
  }
  printf("\n");
}


// Function from question
struct ListNode* removeElements(struct ListNode* head, int val) {

    struct ListNode *beforeNode = head;

    while(beforeNode != NULL){

        if(head == beforeNode && head->val == val){
            struct ListNode* q = head;
            head = q->next;
            beforeNode = head;
            free(q);
        }
        else if(beforeNode->next != NULL && beforeNode->next->val == val){
            struct ListNode *p = beforeNode->next;
            beforeNode = p->next;
            printf("Free val %p %d\n", (void*)p, p->val);
            free(p);
        }

       else
           beforeNode = beforeNode->next;

    }

    return head;
}

int main()
{
  // Initialize a list with three element like: 1->42->1->NULL
  struct ListNode *head = malloc(sizeof *head);
  head->val = 1;
  head->next = malloc(sizeof *head);
  head->next->val = 42;
  head->next->next = malloc(sizeof *head);
  head->next->next->val = 1;
  head->next->next->next = NULL;

  printList(head);

  removeElements(head, 42);

  printList(head);

  return 0;
}

示例输出:

0x558311c02260 1->0x558311c02280 42->0x558311c022a0 1
Free val 0x558311c02280 42
0x558311c02260 1->0x558311c02280 42

如您所见，存在两个问题:

• 结果列表是 1->42，但我们预期是 1->1 换句话说 - 列表已损坏。

• 最后一行打印的节点是我们刚刚释放的节点(即0x558311c02280)。这是未定义的行为。

问题出在这一行:

beforeNode = p->next;

应该是

beforeNode->next = p->next;

上述更改后的输出:

0x561def3e6260 1->0x561def3e6280 42->0x561def3e62a0 1
Free val 0x561def3e6280 42
0x561def3e6260 1->0x561def3e62a0 1

现在列表是正确的，并且没有使用已释放的内存。