Skip to content

Latest commit

 

History

History
101 lines (85 loc) · 2.48 KB

1.1.4 LRU缓存机制.md

File metadata and controls

101 lines (85 loc) · 2.48 KB

题目:LRU 缓存机制 设计和实现一个 LRU(最近最少使用)缓存数据结构,使它应该支持一下操作:get 和 put。 get(key) - 如果 key 存在于缓存中,则获取 key 的 value(总是正数),否则返回 -1。 put(key,value) - 如果 key 不存在,请设置或插入 value。当缓存达到其容量时,它应该在插入新项目之前使最近最少使用的项目作废。

出题人:文景/阿里云 CDN 资深技术专家

参考答案

python版本的:

class LRUCache(object):
    def __init__(self, capacity):
    """
    :type capacity: int
    """
    self.cache = {}
    self.keys = []
    self.capacity = capacity
    
    def visit_key(self, key):
        if key in self.keys:
            self.keys.remove(key)
        self.keys.append(key)
    
    def elim_key(self):
        key = self.keys[0]
        self.keys = self.keys[1:]
        del self.cache[key]
        
    def get(self, key):
        """
        :type key: int
        :rtype: int
        """
        if not key in self.cache:
            return -1
        self.visit_key(key)
        return self.cache[key]
    
    def put(self, key, value):
        """
        :type key: int
        :type value: int
        :rtype: void
        """
        if not key in self.cache:
        if len(self.keys) == self.capacity:
        self.elim_key()
        self.cache[key] = value
        self.visit_key(key)

def main():
    s =
    [["put","put","get","put","get","put","get","get","get"],[[1,1],[2,2],[1],[3,3],[2],[
    4,4],[1],[3],[4]]]
    obj = LRUCache(2)
    l=[]
    for i,c in enumerate(s[0]):
        if(c == "get"):
            l.append(obj.get(s[1][i][0]))
        else:
            obj.put(s[1][i][0], s[1][i][1])
    print(l)

if __name__ == "__main__":
    main()

c++版本的:

class LRUCache{
    public:
        LRUCache(int capacity) {
            cap = capacity;
        }
        
        int get(int key) {
            auto it = m.find(key);
            if (it == m.end()) return -1;
            l.splice(l.begin(), l, it->second);
            return it->second->second;
        }
        
        void set(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();
            if (m.size() > cap) {
                int k = l.rbegin()->first;
                l.pop_back();
                m.erase(k);
            }
        }
}