题目：
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.
解题思路：
每一个Cache行有三个域（key, value, update_cnt)；
用unordered_map来存储Cache体；
用queue来保存使用的历史信息。
代码：
struct Line { int key; int value; int update_cnt; }; class LRUCache{ public: int capacity; unordered_map<int, Line> cache; queue<Line> q; LRUCache(int capacity) { this->capacity = capacity; } int get(int key) { if (cache.count(key)) { cache[key].update_cnt++; q.push(cache[key]); return cache[key].value; } return -1; } void set(int key, int value) { if (cache.count(key)) { cache[key].value = value; cache[key].update_cnt++; q.push(cache[key]); } else { if (cache.size() == capacity) { Line tmp; while (!q.empty()) { tmp = q.front(); q.pop(); if (cache.count(tmp.key) && tmp.update_cnt == cache[tmp.key].update_cnt) { break; } } cache.erase(cache.find(tmp.key)); cache[key].key = key; cache[key].value = value; cache[key].update_cnt = 0; q.push(cache[key]); return; } cache[key].key = key; cache[key].value = value; cache[key].update_cnt = 0; q.push(cache[key]); } } };
LeetCode OJ - LRU Cache,布布扣,bubuko.com
原文：http://www.cnblogs.com/dongguangqing/p/3726319.html