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。 LFU 就是这样一种页面替换策略，其中最不常用的页面被替换。如果页面出现的频率相同，则首先替换最先到达的页面。
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，使用最不常用 (LFU) 算法找出页面错误的数量。
示例-1：
Input : N = 7, C = 3
pages = { 1, 2, 3, 4, 2, 1, 5 }
Output :
Page Faults = 6
Page Hits = 1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 3 is required, since it is not present,
it is a page fault : page fault = 2 + 1 = 3
Page 4 is required, since it is not present,
it replaces LFU page 1 : page fault = 3 + 1 = 4
Page 2 is required, since it is present,
it is not a page fault : page fault = 4 + 0 = 4
Page 1 is required, since it is not present,
it replaces LFU page 3 : page fault = 4 + 1 = 5
Page 5 is required, since it is not present,
it replaces LFU page 4 : page fault = 5 + 1 = 6
示例 2 ：
Input : N = 9, C = 4
pages = { 5, 0, 1, 3, 2, 4, 1, 0, 5 }
Output :
Page Faults = 8
Page Hits = 1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 LFU page 5 : page fault = 4 + 1 = 5
Page 4 is required, since it is not present,
it replaces LFU 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 LFU 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.
Create a map to store frequency of pages
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
increase its frequency
else:
if(memory is full)
remove the first element and decrease frequency of 1st element
Increment count
push the element at the end and increase its frequency
Compare frequency with other pages starting from the 2nd last page
Sort the pages based on their frequency and time at which they arrive
if frequency is same, then, the page arriving first must be placed first 下面是上述算法的实现。
C++
// C++ program to illustrate
// page faults in LFU
#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;
// To store frequency of pages
unordered_map mp;
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) {
// Decrease the frequency
mp[v[0]]--;
// Remove the first element as
// It is least frequently used
v.erase(v.begin());
}
// Add the element at the end of memory
v.push_back(pages[i]);
// Increase its frequency
mp[pages[i]]++;
// Increment the count
count++;
}
else {
// If element is present
// Remove the element
// And add it at the end
// Increase its frequency
mp[pages[i]]++;
v.erase(it);
v.push_back(pages[i]);
}
// Compare frequency with other pages
// starting from the 2nd last page
int k = v.size() - 2;
// Sort the pages based on their frequency
// And time at which they arrive
// if frequency is same
// then, the page arriving first must be placed first
while (mp[v[k]] > mp[v[k + 1]] && k > -1) {
swap(v[k + 1], v[k]);
k--;
}
}
// Return total page faults
return count;
}
/* Driver program to test pageFaults function*/
int main()
{
int pages[] = { 1, 2, 3, 4, 2, 1, 5 };
int n = 7, c = 3;
cout << "Page Faults = " << pageFaults(n, c, pages)
<< endl;
cout << "Page Hits = " << n - pageFaults(n, c, pages);
return 0;
}
// This code is contributed by rajsanghavi9. Page Faults = 6
Page Hits = 1