我已经尝试从LeetCode实现LRU缓存问题。https://leetcode.com/problems/lru-cache/缓存中的条目以{key,value}对的形式存在。我面临着一个奇怪的问题。为了解决这个问题,方法是使用一个带有hashmap的双重链表。对于双链表,我所做的是在我的C++代码中使用了一个deque STL容器而不是list STL容器,但是我的代码在下面的测试用例中失败了。如果我用list替换deque,代码就会被接受。有人可以帮助什么是与使用deque在实现一个LRU缓存的问题。此外,我的代码使用Deque通过了许多测试用例。请帮帮我。
我的代码:
class LRUCache {
public:
unordered_map<int, pair<deque<int>::iterator,int>> mp; // {key, pair<iterator_position_in_deque, value>}; // iterator is the position in the deque where the key is stored.
deque<int> dq; // {key} storing only keys.
int size;
int curr_size;
LRUCache(int capacity) {
curr_size = 0;
size = capacity;
dq.clear();
mp.clear();
}
int get(int key) {
if(mp.find(key) == mp.end()) {
return -1;
}
// below 3 lines are same as in put function
dq.erase(mp[key].first);
dq.push_front(key);
mp[key].first = dq.begin();
return mp[key].second;
}
void put(int key, int value) {
if(mp.find(key) != mp.end()) { // if key present in cache
dq.erase(mp[key].first);
dq.push_front(key);
mp[key].first = dq.begin();
mp[key].second = value; // update the new value to that key
} else { // if key not present in our map and deque afterwards
if (curr_size < size) {
dq.push_front(key);
mp[key] = make_pair(dq.begin(), value);
curr_size++;
} else if (curr_size == size) {
mp.erase(dq.back()); // element will the key as deque stores key
dq.pop_back();
dq.push_front(key);
mp[key] = make_pair(dq.begin(), value);
}
}
}
};
/**
* Your LRUCache object will be instantiated and called as such:
* LRUCache* obj = new LRUCache(capacity);
* int param_1 = obj->get(key);
* obj->put(key,value);
*/
测试失败的测试用例:
["LRUCache","put","put","put","put","put","get","put","get","get","put","get","put","put","put","get","put","get","get","get","get","put","put","get","get","get","put","put","get","put","get","put","get","get","get","put","put","put","get","put","get","get","put","put","get","put","put","put","put","get","put","put","get","put","put","get","put","put","put","put","put","get","put","put","get","put","get","get","get","put","get","get","put","put","put","put","get","put","put","put","put","get","get","get","put","put","put","get","put","put","put","get","put","put","put","get","get","get","put","put","put","put","get","put","put","put","put","put","put","put"]
[[10],[10,13],[3,17],[6,11],[10,5],[9,10],[13],[2,19],[2],[3],[5,25],[8],[9,22],[5,5],[1,30],[11],[9,12],[7],[5],[8],[9],[4,30],[9,3],[9],[10],[10],[6,14],[3,1],[3],[10,11],[8],[2,14],[1],[5],[4],[11,4],[12,24],[5,18],[13],[7,23],[8],[12],[3,27],[2,12],[5],[2,9],[13,4],[8,18],[1,7],[6],[9,29],[8,21],[5],[6,30],[1,12],[10],[4,15],[7,22],[11,26],[8,17],[9,29],[5],[3,4],[11,30],[12],[4,29],[3],[9],[6],[3,4],[1],[10],[3,29],[10,28],[1,20],[11,13],[3],[3,12],[3,8],[10,9],[3,26],[8],[7],[5],[13,17],[2,27],[11,15],[12],[9,19],[2,15],[3,16],[1],[12,17],[9,1],[6,19],[4],[5],[5],[8,1],[11,7],[5,2],[9,28],[1],[2,2],[7,4],[4,22],[7,24],[9,26],[13,28],[11,26]] ```
Output :
```[null,null,null,null,null,null,-1,null,19,17,null,-1,null,null,null,-1,null,-1,5,-1,12,null,null,3,5,5,null,null,1,null,-1,null,30,5,30,null,null,null,-1,null,-1,24,null,null,18,null,null,null,null,-1,null,null,18,null,null,11,null,null,null,null,null,18,null,null,-1,null,4,29,30,null,12,11,null,null,null,null,29,null,null,null,null,17,22,18,null,null,null,-1,null,null,null,-1,null,null,null,29,18,18,null,null,null,null,-1,null,null,null,null,null,null,null]```
Expected Output :
```[null,null,null,null,null,null,-1,null,19,17,null,-1,null,null,null,-1,null,-1,5,-1,12,null,null,3,5,5,null,null,1,null,-1,null,30,5,30,null,null,null,-1,null,-1,24,null,null,18,null,null,null,null,-1,null,null,18,null,null,11,null,null,null,null,null,18,null,null,-1,null,4,29,30,null,12,11,null,null,null,null,29,null,null,null,null,17,22,18,null,null,null,-1,null,null,null,-1,null,null,null,29,18,18,null,null,null,null,-1,null,null,null,null,null,null,null]```
dq.erase(mp[key].first);
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从deque
中间擦除会使出队列的所有其他现有迭代器无效。仅从deque
的开头或结尾进行擦除不会使任何其他迭代器对未擦除的deque
值无效。
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