• [LeetCode] LRU Cache, Solution


    Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and set.

    get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
    set(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.

    [Thoughts]

    首先,对于cache,如果希望有O(1)的查找复杂度,肯定要用hashmap来保存key和对象的映射。对于LRU而言,问题在于如何用O(1)解决cache entry的替换问题。

    简单的说,cache的存储是一个链表的话,那么只要保证从头到尾的顺序就是cache从新到旧的顺序就好了,对于任何一个节点,如果被访问了,那么就将该节点移至头部。如果cache已满,那么就把尾部的删掉,从头部插入新节点。

    所以,需要用到两个数据结构

    1. hashmap, 保存key和对象位置的映射

    2. list,保存对象新旧程度的序列。不一定是list,也可以用vector,不过list的好处是已经实现了头尾操作的api,vector的话,还要自己写,麻烦。

     

    [Code]

    1 class LRUCache{
    2 public:
    3 struct CacheEntry
    4 {
    5 public:
    6 int key;
    7 int value;
    8 CacheEntry(int k, int v) :key(k), value(v) {}
    9 };
    10
    11 LRUCache(int capacity) {
    12 m_capacity = capacity;
    13 }
    14
    15 int get(int key) {
    16 if (m_map.find(key) == m_map.end())
    17 return -1;
    18
    19 MoveToHead(key);
    20 return m_map[key]->value;
    21 }
    22
    23 void set(int key, int value) {
    24 if (m_map.find(key) == m_map.end())
    25 {
    26 CacheEntry newItem(key, value);
    27 if (m_LRU_cache.size() >= m_capacity)
    28 {
    29 //remove from tail
    30 m_map.erase(m_LRU_cache.back().key);
    31 m_LRU_cache.pop_back();
    32 }
    33
    34 // insert in head.
    35 m_LRU_cache.push_front(newItem);
    36 m_map[key] = m_LRU_cache.begin();
    37 return;
    38 }
    39
    40 m_map[key]->value = value;
    41 MoveToHead(key);
    42 }
    43
    44 private:
    45 unordered_map<int, list<CacheEntry>::iterator> m_map;
    46 list<CacheEntry> m_LRU_cache;
    47 int m_capacity;
    48
    49 void MoveToHead(int key)
    50 {
    51 //Move key from current location to head
    52 auto updateEntry = *m_map[key];
    53 m_LRU_cache.erase(m_map[key]);
    54 m_LRU_cache.push_front(updateEntry);
    55 m_map[key] = m_LRU_cache.begin();
    56 }
    57
    58 };
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  • 原文地址:https://www.cnblogs.com/codingtmd/p/5078848.html
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