LRU算法(Least Recently Used) 算是我們經(jīng)常遇到的一種淘汰算法，其中內存管理模塊進行內存頁回收時有用到，針對不經(jīng)常使用的內存頁，LRU淘汰策略能夠將該內存頁回收給操作系統(tǒng)。

屬于 我們操作系統(tǒng)設計中的 時間局部性原理，最長時間未被訪問的數(shù)據(jù)優(yōu)先淘汰，當內存中已存在的數(shù)據(jù)再次被訪問時，則進行熱度的提升。

本文為了鞏固數(shù)據(jù)結構相關知識，特 使用鏈表數(shù)據(jù)結構方式完整實現(xiàn)LRU鏈表的功能。關于數(shù)據(jù)結構和算法相關的學習導圖可以借鑒數(shù)據(jù)結構和算法導圖

實現(xiàn)算法思路：

   LRU 功能的實現(xiàn):1. 已有鏈表中存在 待插入元素，則將該元素在鏈表中刪除，并頭插法到鏈表頭節(jié)點2. 已有鏈表中不存在 待插入元素：a. 若 LRU 容量已滿，則刪除尾元素，頭插當前元素b. 若 LRU 容量未滿，則直接頭插法當前元素 

以上實現(xiàn) 使用其他的線性表數(shù)據(jù)結構（棧、隊列、數(shù)組）均能夠實現(xiàn)，鏈表的實現(xiàn)方式如下：

#include <iostream>using namespace std;typedef int DataType;/*節(jié)點類*/
class Node{public:DataType val;Node *next;
};class List{
private:int maxSize;int length;Node *head;public:List();List(int size);~List();void insertElementBegin(DataType x); //頭插法建立鏈表bool findElement(DataType x);        //查找鏈表中是否存在元素x,存在則返回true，不存在則返回falsevoid deleteElementEnd();             //刪除鏈表的最后一個節(jié)點bool deleteElem(DataType x);         //刪除鏈表中值為x的節(jié)點，刪除成功則返回true,不存在則返回falsebool isEmpty();                      //判斷鏈表是否為空，是則返回true,否則返回falsebool isFull();                       //判斷鏈表是否滿，是則返回true,否則返回falsevoid printAll();                     //打印鏈表中的元素，鏈表長度，LRU長度void * findElemOptim(DataType x);       //針對此應用的優(yōu)化，查找,返回指定元素的前一個節(jié)點的指針void deleteElemOptim(void * node);     //針對此應用的優(yōu)化，刪除
};List::List(){head = new Node;head -> next = NULL;this -> length = 0;this -> maxSize = 10;
}List::List(int size) {head = new Node;head -> next = NULL;this -> length = 0;this -> maxSize = size;
}List::~List() {Node* p, *tmp;p = head;while(p -> next) {tmp = p -> next;p -> next = p -> next -> next;delete tmp;}delete head;this -> head = NULL;this -> length = 0;
}void List::insertElementBegin(DataType x) {Node *p = new Node;p -> val = x;p -> next = head -> next;head -> next = p;this -> length ++;
}bool List::findElement(DataType x) {Node *p = head;while(p -> next != NULL) {if(p -> val == x) {return true;}p = p -> next;}return false;
}void List::deleteElementEnd() {if(head -> next == NULL) {return;}Node *tmp;Node *p = head;while(p -> next != NULL && p -> next -> next != NULL) {p = p -> next;}tmp = p -> next;p -> next = tmp -> next;this -> length --;delete tmp;
}bool List::deleteElem(DataType x) {Node *p = head;Node *tmp;while(p -> next != NULL) {if(p -> next -> val == x) {tmp = p -> next;p -> next = tmp -> next;delete tmp;this -> length --;return true;}p = p -> next;}return false;
}bool List::isEmpty() {return this -> length == 0 ? true:false;
}bool List::isFull() {return this -> length == maxSize ? true:false;
}void List::printAll() {Node *p;p = head;cout << "lru list length :" << this -> length << endl;cout << "lru list maxSize :" << this -> maxSize << endl;cout << "lru list val:" << endl;while(p -> next != NULL) {p = p -> next;cout << p -> val << " ";}cout << endl;
}void * List::findElemOptim(DataType x) {Node *p = head;while(p -> next != NULL) {if(p -> next -> val == x) {return (void *)p;}p = p -> next;}return NULL;
}void List::deleteElemOptim(void *node) {Node *p, *tmp;p = (Node*)node;tmp = p -> next;p -> next = tmp -> next;delete tmp;this -> length --;
}int main() {cout << "test LRU:" << endl;List list(10);int num = 0;while(1) {cout << "please enter a number, 99999 = exit" << endl;cin >> num;if(num == 9999) {break;}/*LRU 功能的實現(xiàn):1. 已有鏈表中存在 待插入元素，則將該元素在鏈表中刪除，并頭插法到鏈表頭節(jié)點2. 已有鏈表中不存在 待插入元素：a. 若 LRU 容量已滿，則刪除尾元素，頭插當前元素b. 若 LRU 容量未滿，則直接頭插法當前元素           */Node *node = (Node*)list.findElemOptim(num);if(node != NULL) {list.deleteElemOptim(node);list.insertElementBegin(num);} else {if(list.isFull()) {list.deleteElementEnd();list.insertElementBegin(num);} else {list.insertElementBegin(num);}}list.printAll();}return 0;
}

編譯g++ -std=c++11 lru_list.cc

運行如下./a.out：

test LRU:
please enter a number, 99999 = exit
2
lru list length :1
lru list maxSize :10
lru list val:
2 
please enter a number, 99999 = exit
3
lru list length :2
lru list maxSize :10
lru list val:
3 2 
please enter a number, 99999 = exit
4
lru list length :3
lru list maxSize :10
lru list val:
4 3 2
...
...
...
please enter a number, 99999 = exit #輸入了34，此時LRU容量已滿，則刪除尾元素，頭插入34
34
lru list length :10
lru list maxSize :10
lru list val:
34 23 12 11 1 8 7 3 2 6 
please enter a number, 99999 = exit #輸入23，提升23的熱度
23
lru list length :10
lru list maxSize :10
lru list val:
23 34 12 11 1 8 7 3 2 6 
please enter a number, 99999 = exit #輸入2，提升2的熱度
2
lru list length :10
lru list maxSize :10
lru list val:
2 23 34 12 11 1 8 7 3 6 
please enter a number, 99999 = exit
9999

總結

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