给定一组具有不同时间要求的作业，其中每个作业由开始时间、结束时间和CPU 负载组成。如果所有作业都在同一台机器上运行，则任务是随时找到最大 CPU 负载。
例子：
Input: jobs[] = {{1, 4, 3}, {2, 5, 4}, {7, 9, 6}}
Output: 7
Explanation:
In the above-given jobs, there are two jobs which overlaps.
That is, Job [1, 4, 3] and [2, 5, 4] overlaps for the time period in [2, 4]
Hence, the maximum CPU Load at this instant will be maximum (3 + 4 = 7).
Input: jobs[] = {{6, 7, 10}, {2, 4, 11}, {8, 12, 15}}
Output: 15
Explanation:
Since, There are no jobs that overlaps.
Maximum CPU Load will be – max(10, 11, 15) = 15
这个问题一般是Merge Intervals的应用。
方法：这个想法是根据作业的结束时间为作业维护最小堆。然后，为每个实例找到完成的作业并将它们从最小堆中删除。即检查最小堆中作业的结束时间是否在当前作业的开始时间之前结束。同样在每个实例中，通过获取最小堆中存在的所有作业的总和，找到机器上的最大 CPU 负载。
例如：
Given Jobs be {{1, 4, 3}, {2, 5, 4}, {7, 9, 6}}
Min-Heap - {}
Instance 1:
The job {1, 4, 3} is inserted into the min-heap
Min-Heap - {{1, 4, 3}},
Total CPU Load = 3
Instance 2:
The job {2, 5, 4} is inserted into the min-heap.
While the job {1, 4, 3} is still in the CPU,
because end-time of Job 1 is greater than
the start time of the new job {2, 5, 4}.
Min-Heap - {{1, 4, 3}, {2, 5, 4}}
Total CPU Load = 4 + 3 = 7
Instance 3:
The job {7, 9, 6} is inserted into the min-heap.
After popping up all the other jobs because their
end time is less than the start time of the new job
Min Heap - {7, 9, 6}
Total CPU Load = 6
Maximum CPU Load = max(3, 7, 6) = 7下面是上述方法的实现：
C++
// C++ implementation to find the
// maximum CPU Load from the given
// lists of the jobs
#include
#include
#include
#include
using namespace std;
// Blueprint of the job
class Job {
public:
int start = 0;
int end = 0;
int cpuLoad = 0;
// Constructor function for
// the CPU Job
Job(int start, int end,
int cpuLoad)
{
this->start = start;
this->end = end;
this->cpuLoad = cpuLoad;
}
};
class MaximumCPULoad {
// Structure to compare two
// CPU Jobs by their end time
public:
struct endCompare {
bool operator()(const Job& x,
const Job& y) {
return x.end > y.end;
}
};
// Function to find the maximum
// CPU Load at any instance of
// the time for given jobs
static int findMaxCPULoad(
vector& jobs)
{
// Condition when there are
// no jobs then CPU Load is 0
if (jobs.empty()) {
return 0;
}
// Sorting all the jobs
// by their start time
sort(jobs.begin(), jobs.end(),
[](const Job& a, const Job& b) {
return a.start < b.start;
});
int maxCPULoad = 0;
int currentCPULoad = 0;
// Min-heap implemented using the
// help of the priority queue
priority_queue,
endCompare> minHeap;
// Loop to iterate over all the
// jobs from the given list
for (auto job : jobs) {
// Loop to remove all jobs from
// the heap which is ended
while (!minHeap.empty() &&
job.start > minHeap.top().end) {
currentCPULoad -=
minHeap.top().cpuLoad;
minHeap.pop();
}
// Add the current Job to CPU
minHeap.push(job);
currentCPULoad += job.cpuLoad;
maxCPULoad = max(maxCPULoad,
currentCPULoad);
}
return maxCPULoad;
}
};
// Driver Code
int main(int argc, char* argv[])
{
vector input = { { 1, 4, 3 },
{ 7, 9, 6 }, { 2, 5, 4 } };
cout << "Maximum CPU load at any time: "
<< MaximumCPULoad::findMaxCPULoad(input)
<< endl;
} Python3
# Python implementation to find the
# maximum CPU Load from the given
# lists of the jobs
from heapq import *
# Blueprint of the job
class job:
# Constructor of the Job
def __init__(self, start,\
end, cpu_load):
self.start = start
self.end = end
self.cpu_load = cpu_load
# Operator overloading for the
# Object that is Job
def __lt__(self, other):
# min heap based on job.end
return self.end < other.end
# Function to find the maximum
# CPU Load for the given list of jobs
def find_max_cpu_load(jobs):
# Sort the jobs by start time
jobs.sort(key = lambda x: x.start)
max_cpu_load, current_cpu_load = 0, 0
# Min-Heap
min_heap = []
# Loop to iterate over the list
# of the jobs given for the CPU
for j in jobs:
# Remove all the jobs from
# the min-heap which ended
while(len(min_heap) > 0 and\
j.start >= min_heap[0].end):
current_cpu_load -= min_heap[0].cpu_load
heappop(min_heap)
# Add the current job
# into min_heap
heappush(min_heap, j)
current_cpu_load += j.cpu_load
max_cpu_load = max(max_cpu_load,
current_cpu_load)
return max_cpu_load
# Driver Code
if __name__ == "__main__":
jobs = [job(1, 4, 3), job(2, 5, 4),\
job(7, 9, 6)]
print("Maximum CPU load at any time: " +\
str(find_max_cpu_load(jobs)))Java
// JAVA Implementation of the above
// approach
import java.util.Arrays;
import java.util.*;
public class MaximumCpuLoad {
private static int maxCpuLoad(int[][] process) {
Arrays.sort(process,(a,b)->'
{
return a[0]-b[0];
});
// list of intervals
List li = new LinkedList();
// variable to store the result
int ans=0;
// Merge intervals
for(int[] p : process)
{
if(!li.isEmpty() && p[0] 输出
Maximum CPU load at any time: 7性能分析：
- 时间复杂度： O(N*logN)
- 辅助空间： O(N)
方法 2 – 不使用堆，合并重叠的区间。
这也可以通过使用合并间隔的想法来解决。
这个想法相当简单——> 合并重叠的间隔并添加它们的负载。
下面是相同的Java代码。
Java
// JAVA Implementation of the above
// approach
import java.util.Arrays;
import java.util.*;
public class MaximumCpuLoad {
private static int maxCpuLoad(int[][] process) {
Arrays.sort(process,(a,b)->'
{
return a[0]-b[0];
});
// list of intervals
List li = new LinkedList();
// variable to store the result
int ans=0;
// Merge intervals
for(int[] p : process)
{
if(!li.isEmpty() && p[0] 输出
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