leetcode 146. LRU Cache

https://leetcode.com/problems/lru-cache/submissions/

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

感觉这题考的不是数据结构 考的是c++

list<pair<int,int>> 是一个双向链表，存储键值(key,value)

map<int,list<pair<int,int>>::iterator>>是用来存放key值所对应的list中的位置

1.get：判断当前的map中 是否有这个key，为啥不是查list呢？因为list不支持这种操作

如果找到：获取到原来迭代器位置再获取到value，将旧的删除，插入开头新的

2.put：

还是从m中找 看之前是否有，

1.有：挪位置 同put，没区别  在这种情况中 it->second 等价于m[key] 

2.没有 长度也没有超限制：直接加入就好，区别在于 it找不到 所以不能用 it->second 要用m[key] 

3.没有 长度超过限制：删除最末尾的，使用迭代器找到end（end是空的）迭代器--是要删除的,注意删除的顺序。先删除map 再删除list

后面同上

class LRUCache {
public:
    int n;
    list<pair<int, int> > l;
    unordered_map<int, list<pair<int, int>>::iterator> m;
    LRUCache(int capacity) {
        n = capacity;
    }
    
    int get(int key) {
        auto it = m.find(key);
        int ans = -1;
        if (it != m.end()) {
            ans = it->second->second;
            l.erase(it->second);
            l.push_front(make_pair(key, ans));
            it->second = l.begin();
        }
        return ans;
    }
    
    void put(int key, int value) {
        auto it = m.find(key);
        if (it != m.end()) {
            l.erase(it->second);
            l.push_front(make_pair(key, value));
            m[key] = l.begin();
        } else if (m.size() < n) {
            l.push_front(make_pair(key, value));
            m[key] = l.begin();
        } else {
            auto it = l.end();
            it --;
            m.erase(it->first);
            l.erase(it);
            l.push_front(make_pair(key, value));
            m[key] = l.begin();
        }
    }
};

/**
 * 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);
 */