---
tags: linux2022
---

# 2022q1 Homework1 (quiz1)
contributed by < [Eddielin0926](https://github.com/Eddielin0926) >

> [quiz1 題目](https://hackmd.io/@sysprog/linux2022-quiz1)

## 測驗 1

參考 LeetCode 編號 1 的題目 Two Sum，以 hash table 的方式求解。

### 結構體定義

`hlist` 用於 hash table 的實作，它的資料結構定義在 [include/linux/types.h](https://github.com/torvalds/linux/blob/master/include/linux/types.h) 中。
採用 double link list 的方式，處理 [hash collision](https://en.wikipedia.org/wiki/Hash_collision) 的資料。

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;


`map_t` 中的 `bits` 用來表示 hash table 的長度，假設 hash 的值是 4 個位元的數值，那麼就有 $2^4$ 種可能，也就是 `1 << 4`， hash table 的長度為 16。

#define MAP_HASH_SIZE(bits) (1 << bits)

### Hash Table 初始化

根據 `bits` 產生長度是 $2^{bits}$ 的 `hlist_head` 的陣列。

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;
}

### Hash

#### Hash Key

double link list 的節點，包了雜湊值以及原本的資料。

struct hash_key {
    int key;
    void *data;
    struct hlist_node node;
};

#### container_of

利用 `offsetof` 反向找到包含這個屬性 (attribute) 的結構體 (struct) 位址。

#define container_of(ptr, type, member)               \
    ({                                                \
        void *__mptr = (void *) (ptr);                \
        ((type *) (__mptr - offsetof(type, member))); \
    })

#### Hash Value

乘上 `GOLDEN_RATIO_32` 後，只取高位的位元，因為高位的位元經過乘法計算後受到的影響必較大，相對來說會比較隨機。

#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);
}

將 `key` 做雜湊後，遍歷 hash table 來找對應的雜湊值。

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;
}

### Map

從 hash table 找對應的資料。

void *map_get(map_t *map, int key)
{
    struct hash_key *kn = find_key(map, key);
    return kn ? kn->data : NULL;
}

當有新的資料進來，如果有找到 key 就直接回傳。
如果沒有找到就要插入到 hash table 中。
`AAA` 和 `BBB` 的答案其實就是要將節點插入到 list 的開頭。

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;
    n->next = first; // AAA
    if (first)
        first->pprev = &n->next;
    h->first = n;
    n->pprev = &h->first; // BBB
}

`deinit` 就要要將 `map` 的記憶體空間釋放出來。

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);
}

整個 twoSum 的概念就是用 `taget - nums[i]` 做 hash table，每個 `nums[i]` 再去找是否有對應的 pair，如果有就會是答案。

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;
}

## 測驗 2

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 (head->next && head->val == head->next->val) { // COND1
        /* Remove all duplicate numbers */
        while (head->next && head->val == head->next->val) // COND2
            head = head->next;
        return deleteDuplicates(head->next);
    }

    head->next = deleteDuplicates(head->next);
    return head;
}

| Question | Answer |
|:-------- |:-------------------------------------------- |
| COND1 | `head->next && head->val == head->next->val` |
| COND2 | `head->next && head->val == head->next->val` |

這邊的做法簡略了釋放記憶體空間的部分，只保留了移除節點的功能。當目前的節點與下一個節點的值相同，就尋找下一個值不相同得節點，並將兩個節點連接起來。

### 避免遞迴

用 `for` 迴圈取代遞迴的動作，利用指標的指標直接修改 `next` 所指向的下一個節點。

#include <stddef.h>

struct ListNode {
    int val;
    struct ListNode *next;
};

struct ListNode *deleteDuplicates(struct ListNode *head)
{
    if (!head)
        return NULL;
    
    for (struct ListNode **ptr = &head; *ptr != NULL; ptr = &((*ptr)->next))
        while ((*ptr)->next && (*ptr)->val == (*ptr)->next->val)
            *ptr = (*ptr)->next;

    return head;
}

### circular doubly-linked list 改寫

#### 遞迴

因為遞迴結束的條件是回到 `head`，因此我多了一層函式來處理 `head`。

#include <stddef.h>

struct ListNode
{
    int val;
    struct ListNode *next;
    struct ListNode *prev;
};

struct ListNode *__deleteDuplicates(struct ListNode *head, struct ListNode *node)
{
    if (!head || !node)
        return NULL;

    if (node == head)
        return node;

    /* Remove all duplicate numbers */
    while (node->next != head && node->val == node->next->val)
        node = node->next->next;

    struct ListNode *next = __deleteDuplicates(head, node->next);
    node->next = next;
    next->prev = node;

    return node;
}

void deleteDuplicates(struct ListNode *head) {
    struct ListNode *next = __deleteDuplicates(head, head->next);
    head->next = next;
    next->prev = head;
}

#### 迭代

與 singly linked list 作法類似，但這邊就不用指標的指標了。

#include <stddef.h>

struct ListNode
{
    int val;
    struct ListNode *next;
    struct ListNode *prev;
};

struct ListNode *deleteDuplicates(struct ListNode *head)
{
    if (!head)
        return NULL;

    for (struct ListNode *ptr = head->next; ptr != head; ptr = ptr->next)
    {
        while (ptr->next != head && ptr->val == ptr->next->val)
        {
            ptr->next->next->prev = ptr;
            ptr->next = ptr->next->next;
        }
    }

    return head;
}

## 測驗 3

針對 LeetCode [146. LRU Cache](https://leetcode.com/problems/lru-cache/)，以下是 [Least Recently Used](https://en.wikipedia.org/wiki/Cache_replacement_policies#Least_recently_used_(LRU)) (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;
    list_for_each_entry_safe (lru, n, &obj->dhead, dlink) { // MMM1
        list_del(&lru->dlink);
        free(lru);
    }
    free(obj); 
}   

int lRUCacheGet(LRUCache *obj, int key)
{
    LRUNode *lru;
    int hash = key % obj->capacity;
    list_for_each_entry (lru, &obj->hheads[hash], hlink) { // MMM2
        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;
    list_for_each_entry (lru, &obj->hheads[hash], hlink) { // MMM3
        if (lru->key == key) {
            list_move(&lru->dlink, &obj->dhead);
            lru->value = value;
            return;
        }
    }

    if (obj->count == obj->capacity) {
        lru = list_entry(&obj->dhead, LRUNode, dlink); // MMM4
        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;
}

| Question | Answer |
|:-------- |:------------------------ |
| MMM1 | `list_for_each_entry_safe` |
| MMM2 | `list_for_each_entry` |
| MMMM3 | `list_for_each_entry` |
| MMM4 | `list_entry` |

### 解釋上述程式碼的運作

### 在 Linux 核心找出 LRU 相關程式碼並探討

[linux/include/linux/lru_cache.h](https://github.com/torvalds/linux/blob/9b76d71fa8be8c52dbc855ab516754f0c93e2980/include/linux/lru_cache.h) 和 [linux/lib/lru_cache.c](https://github.com/torvalds/linux/blob/master/lib/lru_cache.c)

## 測驗 4

#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;
}