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2022q1 Homework1 (quiz1)

contributed by < huang-me >

twoSum

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

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

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;
}

hlist_head 結構:







%0



hlist

hlist_head

first



hlist_node1

hlist_node

pprev

next



hlist:first->hlist_node1:n





hlist_node1:pp->hlist:first





hlist_node2

hlist_node

pprev

next



hlist_node1:next->hlist_node2:n2





hlist_node2:pp2->hlist_node1:next





NULL

NULL



hlist_node2:next2->NULL





  • 在 linked list 中使用指標的指標(pointer to pointer)指向一個指向自己的 pointer 使得我們即使不知道前一個 node 的確切位置依舊可以對 linked list 做插入或者刪除的動作。

Remove Duplicates from Sorted List II

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

    head->next = deleteDuplicates(head->next);
    return head;
}
  • head 之後有其他 node 之值與之相同則將 head 指向下一個值不相同的 node ,反之則對 head->nextdeleteDuplicates

LRUCache

#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) {
        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) {
        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) {
        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);
        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;
}
  • 利用 hash table 使得查詢的平均時間複雜度為 O(1) ,並且利用 linux/list.h 中定義的 macro/function 輔助完成。
  • 其中 container_of 利用 struct 中的子結構位置回推整個母結構的起始位置。

Longest Consecutive Sequence

#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(--left, n_size, heads))) {
                len++;
                list_del(&node->link);
            }

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

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

    return length;
}
  • 因為要從目前的數值 num 開始往左右尋找,而 left、right 的初值與 num 相同,因此得知在執行尋找前需先進行 ++/