leetcode做题笔记146. LRU 缓存-Toy模板网

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请你设计并实现一个满足 LRU (最近最少使用) 缓存约束的数据结构。

实现 LRUCache 类：

LRUCache(int capacity) 以 正整数 作为容量 capacity 初始化 LRU 缓存
int get(int key) 如果关键字 key 存在于缓存中，则返回关键字的值，否则返回 -1 。
void put(int key, int value) 如果关键字 key 已经存在，则变更其数据值 value ；如果不存在，则向缓存中插入该组 key-value 。如果插入操作导致关键字数量超过 capacity ，则应该逐出最久未使用的关键字。

函数 get 和 put 必须以 O(1) 的平均时间复杂度运行。

思路一：双向链表

c语言解法

struct LRUInfo{
    int val;
    int value;
    struct LRUInfo* pre;
    struct LRUInfo* next;
};

typedef struct {
    int top;
    int total;
    struct LRUInfo * head;
    struct LRUInfo * rear;
    struct LRUInfo lruinfo[10001];
} LRUCache;

LRUCache* lRUCacheCreate(int capacity) {
    LRUCache* obj = calloc(1, sizeof(LRUCache));
    obj->total = capacity;
    obj->head = calloc(1, sizeof(struct LRUInfo));
    obj->rear = calloc(1, sizeof(struct LRUInfo));
    obj->head->next = obj->rear;
    obj->rear->pre = obj->head;
    return obj;
}

int lRUCacheGet(LRUCache* obj, int key) {
    if (obj->lruinfo[key].val == 1) {
        obj->lruinfo[key].pre->next = obj->lruinfo[key].next;
        obj->lruinfo[key].next->pre = obj->lruinfo[key].pre;
        obj->rear->pre->next = obj->lruinfo + key;
        obj->lruinfo[key].pre = obj->rear->pre;
        obj->lruinfo[key].next = obj->rear;
        obj->rear->pre = obj->lruinfo + key;
        return obj->lruinfo[key].value;
    }
    return -1;
}

void lRUCachePut(LRUCache* obj, int key, int value) {
    if (obj->lruinfo[key].val == 0 && obj->top < obj->total) {
        (obj->top)++;
        obj->rear->pre->next = obj->lruinfo + key;
        obj->lruinfo[key].pre = obj->rear->pre;
        obj->lruinfo[key].next = obj->rear;
        obj->lruinfo[key].val = 1;
        obj->lruinfo[key].value = value;
        obj->rear->pre = obj->lruinfo + key;
    } else if (obj->lruinfo[key].val == 1){
        obj->lruinfo[key].pre->next = obj->lruinfo[key].next;
        obj->lruinfo[key].next->pre = obj->lruinfo[key].pre;
        obj->rear->pre->next = obj->lruinfo + key;
        obj->lruinfo[key].pre = obj->rear->pre;
        obj->lruinfo[key].next = obj->rear;
        obj->lruinfo[key].value = value;
        obj->rear->pre = obj->lruinfo + key;
    } else if (obj->lruinfo[key].val == 0 && obj->top >= obj->total && obj->head->next != NULL) {
        obj->lruinfo[key].val = 1;
        obj->lruinfo[key].value = value;
        obj->rear->pre->next = obj->lruinfo + key;
        obj->lruinfo[key].pre = obj->rear->pre;
        obj->lruinfo[key].next = obj->rear;
        obj->rear->pre = obj->lruinfo + key;
        obj->head->next->val = 0;
        obj->head->next = obj->head->next->next;
        obj->head->next->pre = obj->head;       
    }
    return;
}

void lRUCacheFree(LRUCache* obj) {
    free(obj);
}

/**
 * Your LRUCache struct will be instantiated and called as such:
 * LRUCache* obj = lRUCacheCreate(capacity);
 * int param_1 = lRUCacheGet(obj, key);
 
 * lRUCachePut(obj, key, value);
 
 * lRUCacheFree(obj);
*/