146. LRU Cache

标签：oid   ret   structure   next   ready   targe   time   int   read   

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.

The cache is initialized with a positive capacity.

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

class LRUCache {
    private Map<Integer, Integer> map;
    private int capacity;
    public LRUCache(int capacity) {
        this.capacity = capacity;
        map = new LinkedHashMap<Integer, Integer>();
    }
    
    public int get(int key) {
        Integer val = map.get(key);
        if (val == null) return -1;
        map.remove(key);
        map.put(key, val);//确保get后当前key变成最后一个插入的
        return val;
    }
    
    public void put(int key, int value) {
        map.remove(key);//Why? Because假如update键的新value也算作visited，要先把当前key删掉重新insert确保是最后一个插入的
        map.put(key, value);
        if (map.size() > capacity){
            // for(Map.Entry<Integer, Integer> entry: map.entrySet()){
            //     System.out.println("key"+entry.getKey()+"value"+entry.getValue());
            // }
            map.remove(map.entrySet().iterator().next().getKey());
        }
    }
}

/**
 * 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);
 */

来自小土刀大神，短小精悍，用了LinkedHashMap的特性（保留插入顺序）。

get的时候先取出value，然后重新插入确保这个key是最后一个值。

put的时候也是要先remove了再重新put进去保证update键的情况

146. LRU Cache

标签：oid   ret   structure   next   ready   targe   time   int   read   

原文地址：https://www.cnblogs.com/wentiliangkaihua/p/11756743.html