题目:
LRU 缓存机制 设计和实现一个 LRU(最近最少使用)缓存数据结构,使它应该支持一下操作:get 和 put。
- get(key) - 如果 key 存在于缓存中,则获取 key 的 value(总是正数),否则返回 -1。
- put(key,value) - 如果 key 不存在,请设置或插入 value。当缓存达到其容量时,它应该在插入新项目之前使最近最少使用的项目作废。
#include <unordered_map>
struct DListNode{
int key, val;
DListNode* pre;
DListNode* next;
DListNode(int k, int v): key(k), val(v), pre(nullptr), next(nullptr){
};
};
class Solution {
private:
int size = 0;
DListNode* head;
DListNode* tail;
unordered_map<int, DListNode*> mp;
public:
/**
* lru design
* @param operators int整型vector<vector<>> the ops
* @param k int整型 the k
* @return int整型vector
*/
vector<int> LRU(vector<vector<int> >& operators, int k) {
// write code here
if(k < 1) return {
};
this->size = k;
this->head = new DListNode(0,0);
this->tail = new DListNode(0,0);
this->head->next = this->tail;
this->tail->pre = this->head;
if(operators.size() == 0) return {
};
vector<int> res;
for(vector<int> op : operators){
if(op[0] == 1) {
set(op[1], op[2]);
}
else if(op[0] == 2){
int value = get(op[1]);
res.push_back(value);
}
}
return res;
}
void set(int key, int val){
if(mp.find(key) == mp.end()){
// hashmap 中没找到
DListNode* node = new DListNode(key, val);
mp[key] = node;
if(this->size <= 0){
removeLast();
}
else{
this->size--;
}
insertFirst(node);
}
else{
// hashmap 中已经有了,也就是链表里也已经有了
mp[key]->val = val;
moveToHead(mp[key]);
}
}
int get(int key){
int ret = -1;
if(mp.find(key) != mp.end()){
ret = mp[key]->val;
moveToHead(mp[key]);
}
return ret;
}
void moveToHead(DListNode* node){
if(node->pre == this->head) return;
node->pre->next = node->next;
node->next->pre = node->pre;
insertFirst(node);
}
void removeLast(){
mp.erase(this->tail->pre->key);
this->tail->pre->pre->next = this->tail; // remove the last node in dlist
this->tail->pre = this->tail->pre->pre;
}
void insertFirst(DListNode* node){
node->pre = this->head;
node->next = this->head->next;
this->head->next->pre = node;
this->head->next = node;
}
};