Problem description:

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:

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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:

Pythonic way is to use OrderedDict()

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class LRUCache:
    def __init__(self, capacity: int):
        self.capacity = capacity
        self.cache = OrderedDict()

    def get(self, key: int) -> int:
        if key in self.cache:
            val = self.cache[key]
            self.cache.move_to_end(key)
            return val
        else:
            return -1

    def put(self, key: int, value: int) -> None:
        if key in self.cache:
            # renew the node
            del self.cache[key]
            self.cache[key] = value
        elif key not in self.cache and len(self.cache) == self.capacity:
            self.cache.popitem(last=False)
            self.cache[key] = value
        else:
            self.cache[key] = value
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class Node:
    def __init__(self, k, v):
        self.key = k
        self.val = v
        self.prev = None
        self.next = None

class LRUCache:
    def __init__(self, capacity):
        self.capacity = capacity
        self.dic = dict()
        self.head = Node(0, 0)
        self.tail = Node(0, 0)
        self.head.next = self.tail
        self.tail.prev = self.head

    def get(self, key):
        if key in self.dic:
            n = self.dic[key]
            self._remove(n)
            self._add(n)
            return n.val
        return -1

    def put(self, key, value):
        if key in self.dic:
            self._remove(self.dic[key])
        n = Node(key, value)
        self._add(n)
        self.dic[key] = n
        if len(self.dic) > self.capacity:
            n = self.head.next
            self._remove(n)
            del self.dic[n.key]

    def _remove(self, node):
        p = node.prev
        n = node.next
        p.next = n
        n.prev = p

    def _add(self, node):
        p = self.tail.prev
        p.next = node
        self.tail.prev = node
        node.prev = p
        node.next = self.tail

Solution: C++

For this question, we need to think how to track a memory block by a key value.
list<pair<int, int>> represents the memory queue, <key, value>
unordered_map<int, list<pair<int, int>>::iterator> can help us find position of memory block in list
size denotes the memory capacity

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class LRUCache {
public:
    //need three variable to store information
    int size;
    list<pair<int, int>> l;
    unordered_map<int, list<pair<int, int>>::iterator> map;
    LRUCache(int capacity) {
        size= capacity;
    }
    
    int get(int key) {
        // find if this key exist in the map
        // if exist, move it to the beginning in the list
        auto it= map.find(key);
        if(it == map.end()) return -1; //can not find, return -1
        l.splice(l.begin(), l, it->second);
        return it->second->second; //the value corresponding to the key is in list
    }
    
    void put(int key, int value) {
        auto it= map.find(key);
        if(it != map.end()) //exist in the map and list, erase and make a new one
            l.erase(it->second);
        
        l.push_front(make_pair(key, value));
        map[key]= l.begin(); 
        
        if(map.size() > size){
            auto index= l.rbegin()->first;
            l.pop_back();
            map.erase(index);
        }   
    }
};

/**
 * 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);
 */

time complexity: $O(1)$
space complexity: $O(n)$
reference:
http://www.cnblogs.com/grandyang/p/4587511.html