2022q1 Homework1 (quiz1)

tags: Linux 核心設計 成大課程

contributed by < hbr890627 >

測驗一

題目

要補完以下的程式碼

#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。

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

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

#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