从指数链表中删除节点 [英] Deleting node from linked list by index
问题描述
这是我的code从链表中删除一个节点。
vec_store 持有 SEQ 和尺寸。变量 SEQ 持有的载体和一个指针。
有关某种原因，否则如果（I＆LT; S-GT＆;大小-1）。不起作用这是最后一个条件
谁能解决这个问题？顺便说一句，这是<强> C code。
无效delete_vec（vec_store S，int i）以
{
如果（ⅰ℃，||仲＆GT;尺寸-1下;ⅰ）
{
的printf（因为索引％d超出范围的\\ n无法删除矢量，我）;
}
否则，如果（我== 0）
{
节点温度;
TEMP = S-＆gt;序列;
S-＆gt;序列= S-＆GT; seq-＆gt;接下来，
S-GT＆; size--;
免费（TEMP）;
}
否则，如果（我== S-＆GT;大小-1）
{
节点temp1中，TEMP2;
// temp1中的malloc =（sizeof运算（节点））;
TEMP2 =的malloc（sizeof运算（节点））;
temp1中= S-＆gt;序列;
如果（temp1-＆gt;接下来== NULL）
{
免费（temp1目录）;
仲＆gt;序列= NULL;
S-GT＆; size--;
输出（S-GT＆;大小-1 \\ n）;
}
其他
{
而（temp1-＆GT;！下次= NULL）
{
TEMP2 = temp1中;
temp1中= temp1-＆gt;接下来，
}
免费（temp1目录）;
temp2-＆gt;接着= NULL;
S-GT＆; size--;
输出（S-GT＆;大小1 2 \\ n）;
}
}
否则，如果（I＆LT; S-GT＆;大小-1）
{
节点temp1中，TEMP2;
诠释J;
temp1中= S-＆gt;序列;
TEMP2 =的malloc（sizeof的（结构node_record））; 为（J = 0; J＆下; = I-1; J ++）
{
TEMP2 = temp1中;
temp1中= temp1-＆gt;接下来，
} 免费（temp1目录）;
temp2-＆gt;接着= NULL;
S-GT＆; size--;
}
}
这code是无法访问的：
如果（temp1-＆gt;接下来== NULL）{
免费（temp1目录）;
仲＆gt;序列= NULL;
S-GT＆; size--;
输出（S-GT＆;大小-1 \\ n）;
}
...因为 temp1目录是列表中的第一个元素，因此，如果第一个元素也是最后一个元素，它只会发生 - 这意味着 S-方式＆gt;大小为1，所以这将是由早期的如果（我== 0）的情况下抓住了
该分配 TEMP2 （发生在两个地方）是假的 - TEMP2 的价值得到反正覆盖，泄漏的内存分配你：
TEMP2 =的malloc（sizeof运算（节点））;
最后，什么可能是你问的谎言这里的问题（在如果（I＆LT; S-＆GT;大小-1）情况下）：
免费（temp1目录）;
temp2-＆gt;接着= NULL;
这剃掉列表中的全部结束。你想保持列表的尾部左右 - 这样的：
temp2-＆gt;接下来= temp1-＆gt;接下来，
免费（temp1目录）;
顺便说一句， TEMP2 和 temp1目录是pretty平淡的变量名 - 什么 previous 和电流还是什么？此外，对于的特殊情况下，如果（我== S-＆GT;大小-1）终止的完全的不必要的 - 它应该由精处理code您对如果（I＆LT; S-GT＆;大小-1）。情况
This is my code for deleting a node from a linked list.
vec_store holds seq and size. Variable seq holds the vectors and a pointer.
For some reason, the else if(i<s->size-1) doesn't work which is the last condition.
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\n",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\n");
}
else
{
while (temp1->next!=NULL)
{
temp2 = temp1;
temp1 = temp1->next;
}
free(temp1);
temp2->next=NULL;
s->size--;
printf("s->size-1 2\n");
}
}
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--;
}
}
This code is unreachable:
if(temp1->next==NULL){
free(temp1);
s->seq=NULL;
s->size--;
printf("s->size-1\n");
}
...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.
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));
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;
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);
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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