题解 | #不重复打印排序数组中相加和为给定值的所有二元组#
设计LRU缓存结构
http://www.nowcoder.com/practice/e3769a5f49894d49b871c09cadd13a61
struct DLinkedNode{
//key 和 value
int key,value;
//双向指针
DLinkedNode* prev;
DLinkedNode* next;
//构造函数
DLinkedNode():key(0),value(0),prev(nullptr),next(nullptr){}
DLinkedNode( int _key, int _value ):key(_key),value(_value),prev(nullptr),next(nullptr){}
};
class Solution {
unordered_map<int, DLinkedNode> cache;
DLinkedNode head;
DLinkedNode* tail;
int size=0;
int capacity=0;
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
capacity = k;
if(capacity<0) return{};
head = new DLinkedNode();
tail = new DLinkedNode();
head->next = tail;
tail->prev = head;
if( !operators.size() )return {};
vector<int> res;</int></int></int>
for( vector<int> op: operators ){
if( op[0]==1 ){//执行set
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 value){
if( !cache.count(key) ){//表中没有,则添加
DLinkedNode* node = new DLinkedNode(key,value);
cache[key] = node;//给哈希表添加key的value
addToHead(node);
if( ++size > capacity ){//超出范围
//移除最后一个节点
DLinkedNode * removed = removeTail();
cache.erase(removed->key);
delete removed;
--size;
}
}else{//表中存在,更新到头部即可
DLinkedNode* node = cache[key];
node->value = value;
moveToHead(node);
}
}
int get(int key){
if(!cache.count(key)) return-1;
//存在key
DLinkedNode* node = cache[key];
moveToHead(node);
return node->value;
}
void addToHead(DLinkedNode* node){
node->prev = head;
node->next = head->next;
head->next->prev = node;
head->next = node;
}
void removeNode(DLinkedNode* node){
node->prev->next = node->next;
node->next->prev = node->prev;
}
void moveToHead(DLinkedNode* node){
removeNode(node);
addToHead(node);
}
DLinkedNode * removeTail(){
DLinkedNode* node = tail->prev;
removeNode(node);
return node;
}};

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