概述 ：
在这里，它使用分页的概念进行内存管理，需要一个页面替换算法来决定当新页面进来时需要替换哪个页面。每当一个新页面被引用并且不在内存中时，页面错误发生，操作系统用新需要的页面替换现有页面之一。 LRU 是一种这样的页面替换策略，其中替换最近最少使用的页面。
例子 ：
给定长度为 N 的 pages[] 数组中的一系列页面，内存容量为 C，使用最近最少使用 (LRU) 算法找出页面错误的数量。
示例-1：
Input : N = 7, C = 3
pages = {1, 2, 1, 4, 2, 3, 5}
Output : 5解释 ：
Capacity is 3, thus, we can store maximum 3 pages at a time.
Page 1 is required, since it is not present,
it is a page fault: page fault = 1
Page 2 is required, since it is not present,
it is a page fault: page fault = 1 + 1 = 2
Page 1 is required, since it is present,
it is not a page fault: page fault = 2 + 0 = 2
Page 4 is required, since it is not present,
it is a page fault: page fault = 2 + 1 = 3
Page 2 is required, since it is present,
it is not a page fault: page fault = 3 + 0 = 3
Page 3 is required, since it is not present,
it replaces LRU page 2: page fault = 3 + 1 = 4
Page 5 is required, since it is not present,
it replaces LRU page 1: page fault = 4 + 1 = 5示例 2 ：
Input : N = 9, C = 4
Pages = {5, 0, 1, 3, 2, 4, 1, 0, 5}
Output : 8解释 ：
Capacity is 4, thus, we can store maximum 4 pages at a time.
Page 5 is required, since it is not present,
it is a page fault: page fault = 1
Page 0 is required, since it is not present,
it is a page fault: page fault = 1 + 1 = 2
Page 1 is required, since it is not present,
it is a page fault: page fault = 2 + 1 = 3
Page 3 is required, since it is not present,
it is a page fault: page fault = 3 + 1 = 4
Page 2 is required, since it is not present,
it replaces LRU page 5: page fault = 4 + 1 = 5
Page 4 is required, since it is not present,
it replaces LRU page 0: page fault = 5 + 1 = 6
Page 1 is required, since it is present,
it is not a page fault: page fault = 6 + 0 = 6
Page 0 is required, since it is not present,
it replaces LRU page 3: page fault = 6 + 1 = 7
Page 5 is required, since it is not present,
it replaces LRU page 2: page fault = 7 + 1 = 8算法 ：
step-1 : Initialize count as 0.
step-2 : Create a vector / array of size equal to memory capacity.
step-3 : Traverse elements of pages[]
step-4 : In each traversal:
if(element is present in memory):
remove the element and push the element at the end
else:
if(memory is full) remove the first element
Increment count
push the element at the end 算法的实现：
以下是该算法在C++中的实现如下。
C++
// C++ program to illustrate
// page faults in LRU
#include
using namespace std;
/* Counts no. of page faults */
int pageFaults(int n, int c, int pages[])
{
// Initialise count to 0
int count = 0;
// To store elements in memory of size c
vector v;
int i;
for (i = 0; i <= n - 1; i++) {
// Find if element is present in memory or not
auto it = find(v.begin(), v.end(), pages[i]);
// If element is not present
if (it == v.end()) {
// If memory is full
if (v.size() == c) {
// Remove the first element
// As it is least recently used
v.erase(v.begin());
}
// Add the recent element into memory
v.push_back(pages[i]);
// Increment the count
count++;
}
else {
// If element is present
// Remove the element
// And add it at the end as it is
// the most recent element
v.erase(it);
v.push_back(pages[i]);
}
}
// Return total page faults
return count;
}
/* Driver program to test pageFaults function*/
int main()
{
int pages[] = { 1, 2, 1, 4, 2, 3, 5 };
int n = 7, c = 3;
cout << "Page Faults = " << pageFaults(n, c, pages);
return 0;
}
// This code is contributed by rajsanghavi9. 输出 ：
Page Faults = 5