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# 460.LFU-Cache

## 题目描述

Design and implement a data structure for Least Frequently Used (LFU) 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 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.
Note that the number of times an item is used is the number of calls to the get and put functions for that item since it was inserted. This number is set to zero when the item is removed.
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(3); // returns 3.
cache.put(4, 4); // evicts key 1.
cache.get(3); // returns 3
cache.get(4); // returns 4

## 代码

### Approach 1: 三个HashMap

class LFUCache {
int min = -1;
HashMap<Integer,Integer> vals; // Key => value
HashMap<Integer,Integer> counts; // key => count
HashMap<Integer,LinkedHashSet<Integer>> lists; // counts => key, 记录min对应的key
int cap;
public LFUCache(int capacity) {
cap = capacity;
vals = new HashMap<>();
counts = new HashMap<>();
lists = new HashMap<>();
}
public int get(int key) {
if (vals.containsKey(key) == false) {
return -1;
}
// Increase current count in counts
int count = counts.get(key);
counts.put(key, count + 1);
// update the count in lists
lists.get(count).remove(key);
if (count == min && lists.get(count).size() == 0) {
min++;
}
return vals.get(key);
}
public void put(int key, int value) {
if (cap <= 0) return;
if (vals.containsKey(key)) {
// contain key
vals.put(key, value);
get(key);
} else {
// not contain key
if (vals.size() >= cap) {
// pruning
int evit = lists.get(min).iterator().next();
lists.get(min).remove(evit);
vals.remove(evit);
}
vals.put(key, value);
counts.put(key, 1);
min = 1;