Leetcode#146 LRU Cache

原题地址

以前Leetcode的测试数据比较弱，单纯用链表做也能过，现在就不行了，大数据会超时。通常大家都是用map+双向链表做的。

我曾经尝试用C++的list容器来写，后来发现map没法保存list的iterator，总是报错，我也不知道为什么。后来只好手写双向链表，真是痛苦啊，一不小心就会出错。怪不得这道题是hard

代码：

 struct Node {
     int key;
     int val;
     Node *prev;
     Node *next;
     Node() : prev(NULL), next(NULL) {}
     Node(int k, int v) : key(k), val(v), prev(NULL), next(NULL) {}
 };

 class LRUCache {
 public:
     map<int, Node*> index;
     Node *head;
     Node *tail;
     int maxSize;

     LRUCache(int capacity) {
         maxSize = capacity;
         head = new Node();
         tail = new Node();
         head->next = tail;
         tail->prev = head;
     }

     int get(int key) {
         if (index.find(key) == index.end())
             return -;

         Node *node = index[key];
         node->prev->next = node->next;
         node->next->prev = node->prev;
         node->prev = head;
         node->next = head->next;
         head->next->prev = node;
         head->next = node;

         return node->val;
     }

     void set(int key, int value) {
         if (maxSize == )
             return;

         if (index.find(key) != index.end()) {
             Node *node = index[key];
             node->val = value;
             node->prev->next = node->next;
             node->next->prev = node->prev;
             node->prev = head;
             node->next = head->next;
             head->next->prev = node;
             head->next = node;
         }
         else {
             Node *node = new Node(key, value);
             node->next = head->next;
             node->prev = head;
             head->next->prev = node;
             head->next = node;
             index[key] = node;
             if (index.size() > maxSize) {
                 node = tail->prev;
                 tail->prev = node->prev;
                 node->prev->next = tail;
                 index.erase(node->key);
                 delete node;
             }
         }
     }
 };