LeetCode 460. LFU Cache
原题链接在这里：https://leetcode.com/problems/lfu-cache/
题目：
Design and implement a data structure for Least Frequently Used (LFU) cache. It should support the following operations: getand 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 reaches its capacity, it should invalidate the least frequently used item before inserting a new item. For the purpose of this problem, when there is a tie (i.e., two or more keys that have the same frequency), the least recently used key would be evicted.
Follow up:
Could you do both operations in O(1) time complexity?
Example:
LFUCache cache = new LFUCache( 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.get(3); // returns 3. cache.put(4, 4); // evicts key 1. cache.get(1); // returns -1 (not found) cache.get(3); // returns 3 cache.get(4); // returns 4
题解：
去掉least frequently used element, 就需要一个min来maintain到目前最不被利用的元素的利用次数.
用三个map, 一个是维护正常key value pair的HashMap<Integer, Integer> keyVals.
第二个是维护每个key的使用次数.
第三个是维护每个count下对应的key set.
当put第一个元素时, min=1, 对应更新keyVals, keyCounts 和 countKeySets.
get时, key的count要加一, 对应调整keyCounts 和 countKeySets. 若这个key的count恰巧是最少使用次数的最后一个值，那么最少使用次数min++.
在达到capacity后在加新key时利用min来找到least frequently used element, 并对应调整keyVals, keyCounts 和 countKeySets.
Note: corner case capacity <= 0.
countToKeys need LinkedHashSet, because when there is even, evict the oldest one.
Time Complexity: get, O(1). put, O(1).
Space: O(n).
AC Java:
1 public class LFUCache { 2 HashMap<Integer, Integer> keyVals; 3 HashMap<Integer, Integer> keyCounts; 4 HashMap<Integer, LinkedHashSet<Integer>> countKeySets; 5 int capacity; 6 int min; 7 8 public LFUCache(int capacity) { 9 this.capacity = capacity; 10 this.min = -1; 11 keyVals = new HashMap<Integer, Integer>(); 12 keyCounts = new HashMap<Integer, Integer>(); 13 countKeySets = new HashMap<Integer, LinkedHashSet<Integer>>(); 14 countKeySets.put(1, new LinkedHashSet<Integer>()); 15 } 16 17 public int get(int key) { 18 if(!keyVals.containsKey(key)){ 19 return -1; 20 } 21 int count = keyCounts.get(key); 22 keyCounts.put(key, count+1); 23 countKeySets.get(count).remove(key); 24 if(count == min && countKeySets.get(count).size() == 0){ 25 min++; 26 } 27 if(!countKeySets.containsKey(count+1)){ 28 countKeySets.put(count+1, new LinkedHashSet<Integer>()); 29 } 30 countKeySets.get(count+1).add(key); 31 return keyVals.get(key); 32 } 33 34 public void put(int key, int value) { 35 if(capacity <= 0){ 36 return; 37 } 38 39 if(keyVals.containsKey(key)){ 40 keyVals.put(key, value); 41 get(key); 42 return; 43 } 44 if(keyVals.size() >= capacity){ 45 int leastFreq = countKeySets.get(min).iterator().next(); 46 keyVals.remove(leastFreq); 47 keyCounts.remove(leastFreq); 48 countKeySets.get(min).remove(leastFreq); 49 } 50 keyVals.put(key, value); 51 keyCounts.put(key, 1); 52 countKeySets.get(1).add(key); 53 min = 1; 54 } 55 } 56 57 /** 58 * Your LFUCache object will be instantiated and called as such: 59 * LFUCache obj = new LFUCache(capacity); 60 * int param_1 = obj.get(key); 61 * obj.put(key,value); 62 */
