Leetcode 146. LRU Cache

Problem:

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

 

Solution:

　　笔试高频题，我在今年笔试中已经两次碰见该题了。这道题是链表和哈希表的结合。对于这道题有几个基础知识必须了解，一个是链表的迭代器失效问题，和vector不同，链表的删除操作只会使迭代器指向的当前节点失效。第二个是第十行的splice函数，它的原型是void splice (iterator position, list& x, iterator i)，其作用是仅将链表x中的i指向的一个节点拼接到position的位置。接下来就很简单了，put函数通过哈希表找到key所在链表中的迭代器，将其删除，然后在链表开头添加新的pair，并将哈希表重新赋值。如果超出限制范围，则在链表和哈希表中删除最后一个节点。对于get函数，通过哈希表找到结果，然后通过splice函数将该节点移动到链表开头。

Code:

 

class LRUCache {
public:
    LRUCache(int capacity) {
        limit = capacity;
    }
    
    int get(int key) {
        auto it = um.find(key);
        if (it == um.end()) return -1;
        l.splice(l.begin(), l, it->second);
        return it->second->second;
    }
    
    void put(int key, int value) {
        auto iter = um.find(key);
        if(iter != um.end()) 
            l.erase(iter->second);
        l.push_front(make_pair(key,value));
        um[key] = l.begin();
        if(um.size() > limit){
            um.erase(l.rbegin()->first);
            l.pop_back();
        }
    }
private:
    int limit;
    list<pair<int, int>> l;
    unordered_map<int, list<pair<int, int>>::iterator> um;
};

/**
 * 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);
 */