# 2022q1 Homework1 (quiz1)
###### tags: `Linux 核心設計` `成大課程`
contributed by < `hbr890627` >
## 測驗一
### 題目
:::spoiler 要補完以下的程式碼
```c
#include <stddef.h>
#include <stdlib.h>
struct hlist_node { struct hlist_node *next, **pprev; };
struct hlist_head { struct hlist_node *first; };
typedef struct { int bits; struct hlist_head *ht; } map_t;
#define MAP_HASH_SIZE(bits) (1 << bits)
map_t *map_init(int bits) {
map_t *map = malloc(sizeof(map_t));
if (!map)
return NULL;
map->bits = bits;
map->ht = malloc(sizeof(struct hlist_head) * MAP_HASH_SIZE(map->bits));
if (map->ht) {
for (int i = 0; i < MAP_HASH_SIZE(map->bits); i++)
(map->ht)[i].first = NULL;
} else {
free(map);
map = NULL;
}
return map;
}
struct hash_key {
int key;
void *data;
struct hlist_node node;
};
#define container_of(ptr, type, member) \
({ \
void *__mptr = (void *) (ptr); \
((type *) (__mptr - offsetof(type, member))); \
})
#define GOLDEN_RATIO_32 0x61C88647
static inline unsigned int hash(unsigned int val, unsigned int bits) {
/* High bits are more random, so use them. */
return (val * GOLDEN_RATIO_32) >> (32 - bits);
}
static struct hash_key *find_key(map_t *map, int key) {
struct hlist_head *head = &(map->ht)[hash(key, map->bits)];
for (struct hlist_node *p = head->first; p; p = p->next) {
struct hash_key *kn = container_of(p, struct hash_key, node);
if (kn->key == key)
return kn;
}
return NULL;
}
void *map_get(map_t *map, int key)
{
struct hash_key *kn = find_key(map, key);
return kn ? kn->data : NULL;
}
void map_add(map_t *map, int key, void *data)
{
struct hash_key *kn = find_key(map, key);
if (kn)
return;
kn = malloc(sizeof(struct hash_key));
kn->key = key, kn->data = data;
struct hlist_head *h = &map->ht[hash(key, map->bits)];
struct hlist_node *n = &kn->node, *first = h->first;
AAA;
if (first)
first->pprev = &n->next;
h->first = n;
BBB;
}
void map_deinit(map_t *map)
{
if (!map)
return;
for (int i = 0; i < MAP_HASH_SIZE(map->bits); i++) {
struct hlist_head *head = &map->ht[i];
for (struct hlist_node *p = head->first; p;) {
struct hash_key *kn = container_of(p, struct hash_key, node);
struct hlist_node *n = p;
p = p->next;
if (!n->pprev) /* unhashed */
goto bail;
struct hlist_node *next = n->next, **pprev = n->pprev;
*pprev = next;
if (next)
next->pprev = pprev;
n->next = NULL, n->pprev = NULL;
bail:
free(kn->data);
free(kn);
}
}
free(map);
}
int *twoSum(int *nums, int numsSize, int target, int *returnSize)
{
map_t *map = map_init(10);
*returnSize = 0;
int *ret = malloc(sizeof(int) * 2);
if (!ret)
goto bail;
for (int i = 0; i < numsSize; i++) {
int *p = map_get(map, target - nums[i]);
if (p) { /* found */
ret[0] = i, ret[1] = *p;
*returnSize = 2;
break;
}
p = malloc(sizeof(int));
*p = i;
map_add(map, nums[i], p);
}
bail:
map_deinit(map);
return ret;
}
```
### AAA = ?
* `(a) /* no operation */`
* `(b) n->pprev = first`
* `(c) n->next = first`
* `(d) n->pprev = n`
### BBB = ?
* `(a) n->pprev = &h->first`
* `(b) n->next = h`
* `(c) n->next = n`
* `(d) n->next = h->first`
* `(e) n->next = &h->first`
:::
---
要完成此題,需要觀察 `map_add` 的部分。
```c
void map_add(map_t *map, int key, void *data)
{
struct hash_key *kn = find_key(map, key);
if (kn)
return;
kn = malloc(sizeof(struct hash_key));
kn->key = key, kn->data = data;
struct hlist_head *h = &map->ht[hash(key, map->bits)];
struct hlist_node *n = &kn->node, *first = h->first;
AAA;
if (first)
first->pprev = &n->next;
h->first = n;
BBB;
}
```
### 解釋
在 `AAA` 後,判斷 `first` 是否存在,而 `first` 代表的是儲存在這個位置的第一個節點, 若存在,就要將 `first` 的 `pprev` 指向新的節點,之後將新的節點更新為新的 `first` ,最後在執行 `BBB`。
少了 `AAA` 與 `BBB` 會變成以下的樣子
```graphviz
digraph graphname {
rankdir=LR;
node[shape=record]
n[label="new node|{<l>pprev|<r>next}"; pos="1,0"];
h[label="head"; pos="3,12!"]
f[label="first|{<l>pprev|<r>next}", shape=record];
h->n;
f:l->n
}
```
由此可知,缺少的是設定新節點的 `next` 以及 `pprev` 。
`AAA : n->next = first`
`BBB : n->pprev = &h->first`
## 測驗二
### 題目
```c
#include <stddef.h>
struct ListNode {
int val;
struct ListNode *next;
};
struct ListNode *deleteDuplicates(struct ListNode *head)
{
if (!head)
return NULL;
if (COND1) {
/* Remove all duplicate numbers */
while (COND2)
head = head->next;
return deleteDuplicates(head->next);
}
head->next = deleteDuplicates(head->next);
return head;
}
```
COND1 = ?
COND2 = ?
---
### 解釋