leetcode演算法設計第三週作業 LRU快取問題
leetcode演算法設計第三週作業 LRU快取問題: https://leetcode.com/problems/lru-cache/description/ Difficulty:Hard Total Accepted:204.4K Total Submissions:955.9K
Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and put.
get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1. put(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.
Follow up: Could you do both operations in O(1) time complexity?
Example:
LRUCache cache = new LRUCache( 2 /* capacity */ ); cache.put(1, 1); cache.put(2, 2); cache.get(1); // returns 1 cache.put(3, 3); // evicts key 2 cache.get(2); // returns -1 (not found) cache.put(4, 4); // evicts key 1 cache.get(1); // returns -1 (not found) cache.get(3); // returns 3 cache.get(4); // returns 4
題目看起來不難,有作業系統基礎知識的對這個應該都不陌生。 主要難在優化和對於容器的選擇上,這個關係到程式碼的效率以及關鍵函式能否達到o(1)的時間複雜度。 下面是我的第一版答案,雖然通過了但時間較慢,用了結構體陣列,每次查詢刪除的時間複雜度都為o(n),有點臃腫。
class LRUCache { private: struct PointStruct { int key = INT16_MIN; int value = INT16_MIN; }; int capacity; PointStruct* ps; int currentSize; public: LRUCache(int capacity) { ps = new PointStruct[capacity]; this->capacity = capacity; currentSize = 0; } int get(int key) { for (int i = currentSize-1; i >=0; i--) { if (ps[i].key == key) { PointStruct temp; temp.key = ps[i].key; temp.value = ps[i].value; for (int j = i; j <currentSize - 1; j++) { ps[j] = ps[j + 1]; } ps[currentSize - 1] = temp; return temp.value; } } return -1;//not found } void put(int key, int value) { for (int i = currentSize - 1; i >= 0; i--) { if (ps[i].key == key) { ps[i].value = value; PointStruct temp; temp = ps[i]; for (int j = i; j < currentSize-1; j++) { ps[j] = ps[j + 1]; } ps[currentSize - 1] = temp; return; } } if(currentSize==capacity){ for (int i = 0; i < currentSize - 1; i++) { ps[i] = ps[i + 1]; } ps[capacity - 1].key = key; ps[capacity - 1].value = value; } else if (currentSize > capacity) { cout << "error" << endl; } else { ps[currentSize].key = key; ps[currentSize].value = value; currentSize++; } return; } };
見到dalao們的高分參考後,發現有兩點差距:
- 在程式碼段後面可以加上以下的lambda解析式提高輸入輸出效率,這個方法的關鍵在於將c++中的cin和cout的緩衝區關閉,可以顯著提升效率避免時間和速度的損失。
static const auto io_sync_off = []()
{
// turn off sync
std::ios::sync_with_stdio(false);
// untie in/out streams
std::cin.tie(nullptr);
return nullptr;
}();
- 結合利用list和map:
list <pair<int, int>> l;
unordered_map <int, list <pair<int, int>>::iterator> mp;
- 使用list的l.splice(l.begin(), l, x, next(x));可以將x挪到最前面。
- 使用mp.find()可以在o(1)之內找到對應的值,將兩者結合起來,可以做到查詢和刪除的時候都可以達到o(1)的時間複雜度。 其中,unordered_map用的是雜湊函式對映,而不是map的紅黑樹,在查詢的速度上效率更高。