---
tags: linux kernel
---
# 2022q1 Homework1 (quiz1)
contributed by <`jimmy-liu1021`>
[測驗連結](https://hackmd.io/@sysprog/linux2022-quiz1#%E6%B8%AC%E9%A9%97-1)
[參考筆記](https://hackmd.io/@jim12312321/linux2022-quiz1) contributed by<`jim12312321`>
## 測驗 1
[LeetCode](https://leetcode.com/) 中編號1的題目 [Two Sum](https://leetcode.com/problems/two-sum/)：題意為給定一個陣列 `nums` 和一個目標值 `target`，求找到 `nums` 中的2個元素相加會等於 `target` 的索引值。例如：`nums` = `[2, 7, 11, 15]`, `target` = `9`，相加為9的元素僅有`2`跟`7`，故回傳這兩個元素的索引值 `[0, 1]`

以下是引入 hash table 的實作，學習 Linux 核心程式碼風格:

#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;
}

請補完程式碼。

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`

### 結構分析

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)

- 首先觀察結構 `map_t`，由一個整數 `bits` 以及一個指向結構 `hlist_head` 的指標 `ht` 所組成
- `bits` 是用來決定 hash table 的大小，由 `MAP_HASH_SIZE(bits)` 可得知，會產生 $2^{bits}$ 個 `hlist_head` 大小的 hash table
- `ht` 則是指向 hash table 的指標，在 `map_init()` 中可以看到 `ht` 指向

        map->ht = malloc(sizeof(struct hlist_head) * MAP_HASH_SIZE(map->bits));

### hash table 初始化

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;
}

## 測驗2 ： 實作 LeetCode 82.

針對[LeetCode 82. Remove Duplicates from Sorted List II](https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/) ，以下是可能的合法 C 程式實作:

#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 = ？

- 此題是要移除所有重複的數字，所以在 `COND1` 時，判斷 `head->next` 是否存在，若存在的話是否與 `head` 所存的值相同，故 `COND1` = `head->next && head.val == head->next.val`
- 若找到第一個重複的數字，應該繼續往下檢查第二個是否也重複，故接下來應檢查 `head->next->next && head->next.val == head->next->next.val`。結合 while 迴圈內的行為，將 `head` 指向 `head->next`，故 `COND2` = `head->next && head.val == head->next.val`

## 測驗 3
針對 LeetCode [146. LRU Cache](https://leetcode.com/problems/lru-cache/)，以下是 [Least Recently Used (LRU)](https://en.wikipedia.org/wiki/Cache_replacement_policies#Least_recently_used_(LRU)) 可能的合法 C 程式實作:

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

typedef struct {
    int capacity, count;             
    struct list_head dhead, hheads[];
} LRUCache;
    
typedef struct {
    int key, value;
    struct list_head hlink, dlink;
} LRUNode;

LRUCache *lRUCacheCreate(int capacity)
{   
    LRUCache *obj = malloc(sizeof(*obj) + capacity * sizeof(struct list_head));
    obj->count = 0;
    obj->capacity = capacity;
    INIT_LIST_HEAD(&obj->dhead);
    for (int i = 0; i < capacity; i++)
        INIT_LIST_HEAD(&obj->hheads[i]);
    return obj;
}
    
void lRUCacheFree(LRUCache *obj)
{       
    LRUNode *lru, *n;
    MMM1 (lru, n, &obj->dhead, dlink) {
        list_del(&lru->dlink);
        free(lru);
    }
    free(obj); 
}   

int lRUCacheGet(LRUCache *obj, int key)
{
    LRUNode *lru;
    int hash = key % obj->capacity;
    MMM2 (lru, &obj->hheads[hash], hlink) {
        if (lru->key == key) {
            list_move(&lru->dlink, &obj->dhead);
            return lru->value;
        }
    }
    return -1;
}

void lRUCachePut(LRUCache *obj, int key, int value)
{
    LRUNode *lru;
    int hash = key % obj->capacity;
    MMM3 (lru, &obj->hheads[hash], hlink) {
        if (lru->key == key) {
            list_move(&lru->dlink, &obj->dhead);
            lru->value = value;
            return;
        }
    }

    if (obj->count == obj->capacity) {
        lru = MMM4(&obj->dhead, LRUNode, dlink);
        list_del(&lru->dlink);
        list_del(&lru->hlink);
    } else {
        lru = malloc(sizeof(LRUNode));
        obj->count++;
    }
    lru->key = key;
    list_add(&lru->dlink, &obj->dhead);
    list_add(&lru->hlink, &obj->hheads[hash]);
    lru->value = value;
}

請補完程式碼，注意作答規範:

- MMM1, MMM2, MMM3, MMM4 都是 Linux 核心風格的 list 巨集，以 list_ 開頭
- 不要出現空白
- 儘量寫出最精簡的程式碼，而且答案也只接受符合上述程式碼排版風格的最精簡形式
## 測驗 4
針對 LeetCode [128. Longest Consecutive Sequence](https://leetcode.com/problems/longest-consecutive-sequence/description/)，以下是可能的合法 C 程式實作:

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

struct seq_node {
    int num;
    struct list_head link;
};                                                                                                                                                            
            
static struct seq_node *find(int num, int size, struct list_head *heads)
{   
    struct seq_node *node;
    int hash = num < 0 ? -num % size : num % size;
    list_for_each_entry (node, &heads[hash], link) {
        if (node->num == num)
            return node;
    }
    return NULL;
}

int longestConsecutive(int *nums, int n_size)
{
    int hash, length = 0;
    struct seq_node *node;
    struct list_head *heads = malloc(n_size * sizeof(*heads));

    for (int i = 0; i < n_size; i++)
        INIT_LIST_HEAD(&heads[i]);

    for (int i = 0; i < n_size; i++) {
        if (!find(nums[i], n_size, heads)) {
            hash = nums[i] < 0 ? -nums[i] % n_size : nums[i] % n_size;
            node = malloc(sizeof(*node));
            node->num = nums[i];
            list_add(&node->link, &heads[hash]);
        }
    }

    for (int i = 0; i < n_size; i++) {
        int len = 0;
        int num;
        node = find(nums[i], n_size, heads);
        while (node) {
            len++;
            num = node->num;
            list_del(&node->link);

            int left = num, right = num;
            while ((node = find(LLL, n_size, heads))) {
                len++;
                list_del(&node->link);
            }

            while ((node = find(RRR, n_size, heads))) {
                len++;
                list_del(&node->link);
            }

            length = len > length ? len : length;
        }
    }

    return length;
}

請補完程式碼

LLL = ?

- `(a)` `left`
- `(b)` `left++`
- `(c)` `++left`
- `(d)` `left--`
- `(e)` `--left`

RRR = ?

- `(a)` `right`
- `(b)` `right++`
- `(c)` `++right`
- `(d)` `right--`
- `(e)` `--right`