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LeetCode 146. LRU Cache

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Question

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/);

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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

Explanation

这道题用linked hashmap来解决，既存着值，同时每当get和put的操作产生，就算是调用了，所以要调整位置。

Code

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public class LRUCache {
    private LinkedHashMap<Integer, Integer> map;
    private int cap;
    public LRUCache(int capacity) {
        this.cap = capacity;
        this.map = new LinkedHashMap<Integer, Integer>(capacity);
    }
    public int get(int key) {
        if (!map.containsKey(key)) {
            return -1;
        }
        int val = map.get(key);
        map.remove(key);
        map.put(key,val);
        return val;
    }
    public void put(int key, int value) {
        if (map.containsKey(key)) {
            map.remove(key);
        } else if (map.size() >= cap) {
            map.remove(map.entrySet().iterator().next().getKey());
        }
        map.put(key,value);
    }
}