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Problem solution · C++

LRU Cache

LRU Cache: a C++ solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Hash-based lookup
Source
walkccc LeetCode Solutions
Length
48 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Hash-based lookup

For LRU Cache, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 48 lines of C++ from the credited upstream file 146-2.cpp.
  • The implementation visibly relies on sequence storage, hash lookup.
  • No explicit loop blocks detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeLRU Cache · C++C++
Use this to learn the idea, then write your own version.
struct Node {  int key;  int value;}; class LRUCache { public:  LRUCache(int capacity) : capacity(capacity) {}   int get(int key) {    const auto it = keyToIterator.find(key);    if (it == keyToIterator.cend())      return -1;     const auto& listIt = it->second;    // Move it to the front.    cache.splice(cache.begin(), cache, listIt);    return listIt->value;  }   void put(int key, int value) {    // There's no capacity issue, so just update the value.    if (const auto it = keyToIterator.find(key); it != keyToIterator.cend()) {      const auto& listIt = it->second;      // Move it to the front.      cache.splice(cache.begin(), cache, listIt);      listIt->value = value;      return;    }     // Check the capacity.    if (cache.size() == capacity) {      const Node& lastNode = cache.back();      // That's why we store `key` in `Node`.      keyToIterator.erase(lastNode.key);      cache.pop_back();    }     cache.emplace_front(key, value);    keyToIterator[key] = cache.begin();  }  private:  const int capacity;  list<Node> cache;  unordered_map<int, list<Node>::iterator> keyToIterator;};

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