首页 > 代码库 > [LeetCode]LRU Cache有个问题,求大神解答

[LeetCode]LRU Cache有个问题,求大神解答

题目:

Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and set.

get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
set(key, value) - Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.

这是我的代码:

 1 class LRUCache{ 2 public: 3     int num; 4     int max; 5     list<int> latest_key;         //用于保存使用情况,队头是最久未使用的,队尾是最近使用的 6     unordered_map<int, int> cache;   //用于保存key,value 7  8     LRUCache(int capacity) { 9         num = 0;10         max = capacity;11     }12 13     int get(int key) {14         unordered_map<int, int>::iterator it = cache.find(key);15         list<int>::iterator iter;16         if (it == cache.end())  //如果没有找到key17             return -1;18         else            //如果找到了key,就在对应的最近latest队列里面修改key的位置,先把key所在的位置删除,再把key放到队尾19         {20             iter = latest_key.begin();21             while (*iter != key)22                 iter++;23             latest_key.erase(iter);24             latest_key.push_back(key);25             return it->second;26         }27     }28 29     void set(int key, int value) {30         unordered_map<int, int>::iterator it = cache.find(key);31         list<int>::iterator iter;32         if (it != cache.end())  //如果要插入的已经有key存在,就先在优先队列里面找到key出现的位置,删除,再把key插入队尾33         {34             it->second = value;35             iter = latest_key.begin();36             while (*iter != key)37                 iter++;38             latest_key.erase(iter);39             latest_key.push_back(key);40         }41         else            //如果要插入的不存在42         {43             if (num<max)  //如果不超过cache容量,则直接在cahe中插入,再在队尾添加该key44             {45                 num++;46                 cache.insert(std::pair< int, int >(key, value));47                 latest_key.push_back(key);48             }49             else      //如果cache已经满了,则根据队头元素,在cache删除对应键值,再在队列中删除这个队头,之后,把新要插入的键值插入cache中,把新key插入队尾50             {51                 int latest = latest_key.front();52                 cache.erase(latest);53                 latest_key.pop_front();54                 cache.insert(std::pair< int, int >(key, value));55                 latest_key.push_back(key);56             }57         }58     }59 };

  当我把代码中出现:

1  iter = latest_key.begin();2  while (*iter != key)3     iter++; 

  部分替换为:

1 iter=find(latest_key.begin(),latest_key.end(),key);

  就会报错:

Time Limit Exceeded

Last executed input: 2048,[set(1178,3401),set(903,6060).....

  我大致查了一下find的实现机制,也是遍历啊,按理说这两者效率差不多,为什么替换之后就不能通过?而替换之前能通过,求大神解答!!

  万分感谢!!!