按索引从链表中删除节点 [英] Deleting node from linked list by index

问题描述

这是我从链表中删除节点的代码.

This is my code for deleting a node from a linked list.

vec_store 保存 seq 和 size.变量 seq 保存向量和指针.

vec_store holds seq and size. Variable seq holds the vectors and a pointer.

出于某种原因，else if(i<s->size-1) 不起作用，这是最后一个条件.

For some reason, the else if(i<s->size-1) doesn't work which is the last condition.

谁能解决这个问题?顺便说一下，这是C 代码.

Can anyone solve the problem? By the way this is C code.

void delete_vec(vec_store s, int i)
{
    if (i<0 || s->size-1<i)
    {
        printf("Cannot delete vector because index %d is out of bounds
",i);
    }
    else if (i==0)
    {
        node temp;
        temp = s->seq;
        s->seq = s->seq->next;
        s->size--;
        free(temp);
    }
    else if(i==s->size-1)
    {
        node temp1, temp2;
        //temp1 = malloc(sizeof (node));
        temp2 = malloc(sizeof (node));
        temp1=s->seq;
        if(temp1->next==NULL) 
        {
            free(temp1);
            s->seq=NULL;
            s->size--;
            printf("s->size-1
");
        }
        else
        {
            while (temp1->next!=NULL)
            {
                temp2 = temp1;
                temp1 = temp1->next;    
            }
            free(temp1);
            temp2->next=NULL;
            s->size--;
            printf("s->size-1 2
");
        }
    }
    else if(i<s->size-1)
    {
        node temp1,temp2;
        int j;
        temp1=s->seq;
        temp2 = malloc(sizeof (struct node_record));

        for(j=0;j<=i-1;j++)
        {
            temp2=temp1;
            temp1 = temp1->next;
        }

        free(temp1);
        temp2->next=NULL;
        s->size--;
    }
}

推荐答案

此代码无法访问:

if(temp1->next==NULL){
free(temp1);
s->seq=NULL;
s->size--;
printf("s->size-1
");
}

...因为 temp1 是列表中的第一个元素，所以它只会在第一个元素也是最后一个元素时发生 - 这意味着 s->size 是 1，所以这会被前面的 if (i==0) 情况捕获.

...because temp1 is the first element in the list, and so it would only happen if the first element was also the last element - which implies s->size is 1, so this would have been caught by the earlier if (i==0) case.

这种对 temp2 的分配(出现在两个地方)是虚假的 - temp2 的值无论如何都会被覆盖，从而泄漏您分配的内存:

This allocation to temp2 (which occurs in two places) is bogus - temp2's value gets overwritten anyway, leaking the memory you allocated:

temp2 = malloc(sizeof (node));

最后，您要问的问题可能是什么(在 if(i<s->size-1) 情况下):

Finally, what is probably the problem you're asking about lies here (in the if(i<s->size-1) case):

free(temp1);
temp2->next=NULL;

这会将整个结尾从列表中删除.你想保留列表的尾部 - 像这样:

This chops the entire end off the list. You want to keep the tail of the list around - like this:

temp2->next = temp1->next;
free(temp1);

顺便说一下，temp2 和 temp1 是非常乏味的变量名 - previous 和 current 或某物?此外，if(i==s->size-1) 的特殊情况 完全 是不必要的 - 它应该由您为 if(isize-1) case.

By the way, temp2 and temp1 are pretty uninspiring variable names - what about previous and current or something? Also, the special case for if(i==s->size-1) is completely unnecessary - it should be handled fine by the code you have for the if(i<s->size-1) case.

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