题目描述：

代码实现：这里记录了根据LRU算法原理最直接理解的代码实现。

import java.util.*;

//存储输入内容，记录访问权值
class CounterInfo {
    int key;
    int value;
    int times;//代表key对应的权值，值越小优先级越高
    public CounterInfo(int key, int value) {
        this.key = key;
        this.value = value;
        this.times = 0;
    }
}

public class Solution {
    ArrayList<CounterInfo> intarr;
    int maxsize;
    public Solution(int capacity) {
        // write code here
        intarr = new ArrayList<CounterInfo>(0);
        this.maxsize = capacity;
    }


    //除了key以外的元素都更新times++
    public void refreshTimes(int key) {
        Iterator infoit = intarr.iterator();
        while (infoit.hasNext()) {
            CounterInfo nowinfo = (CounterInfo)infoit.next();
            if (nowinfo.key != key) {
                nowinfo.times++;
            }
        }
    }


    /*
    1.遍历列表看是否存在key，如果存在则返回相应的value，如果不存在返回-1
    2.如果存在目标key，并且目标key对应权值不为0，更新目标key对应的权值为0，
      其他key对应权值都+1
    3.如果存在目标key，但是目标key对应权值为0，列表内所有权值不做改变
    */
    public int get(int key) {
        boolean isfind = false;
        boolean isrefresh = false;
        int resValue = -1;
        //查找intarr中是否存在key
        Iterator infoit = intarr.iterator();
        while (infoit.hasNext()) {
            CounterInfo nowinfo = (CounterInfo)infoit.next();
            if (nowinfo.key == key) {
                isfind = true;
                resValue = nowinfo.value;
                if (nowinfo.times != 0) {
                    isrefresh = true;
                    nowinfo.times = 0;
                } else {
                    isrefresh = false;
                }
            }
        }
        if (isfind) {
            if (isrefresh) {
                //更新其他info的times
                this.refreshTimes(key);
            }
            return resValue;
        } else {
            return -1;
        }
    }


    /*
    1.看是否存在key，如果存在更新key对应的value和权值=0
    2.如果不存在：
        2.1 列表满：选择权值最大的元素，修改key，value，权值=0；其他元素权值+1
        2.2 列表未满：列表添加新的CounterInfo对象；其他元素权值+1
    */
    public void set(int key, int value) {
        //先看是否存在
        boolean isfind = false;
        //查找intarr中是否存在key
        Iterator infoit = intarr.iterator();
        while (infoit.hasNext()) {
            CounterInfo nowinfo = (CounterInfo)infoit.next();
            if (nowinfo.key == key) {
                isfind = true;
                //更新value
                nowinfo.value = value;
                if (nowinfo.times != 0) {
                    nowinfo.times = 0;
                    this.refreshTimes(key);
                }
            }
        }

        if (!isfind) {
            //判断是否达到最大长度
            if (intarr.size() == maxsize) {
                //找到最久未访问的元素，更新key，value，times
                infoit = intarr.iterator();

                //找到最大time
                int maxtime = 0;
                while (infoit.hasNext()) {
                    CounterInfo nowinfo = (CounterInfo)infoit.next();
                    maxtime = maxtime > nowinfo.times ? maxtime : nowinfo.times;
                }

                //根据最大time更新key-value值
                infoit = intarr.iterator();
                while (infoit.hasNext()) {
                    CounterInfo nowinfo = (CounterInfo)infoit.next();
                    if (nowinfo.times == maxtime) {
                        nowinfo.times = 0;
                        nowinfo.key = key;
                        nowinfo.value = value;
                    }
                }
            } else {
                CounterInfo newinfo = new CounterInfo(key, value);
                intarr.add(newinfo);
            }
            this.refreshTimes(key);
        }

    }
}

/**
 * Your Solution object will be instantiated and called as such:
 * Solution solution = new Solution(capacity);
 * int output = solution.get(key);
 * solution.set(key,value);
 */

刷题链接：

设计LRU缓存结构_牛客题霸_牛客网