leetcode做题笔记146. LRU 缓存

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);
*/