我正在尝试使用 javascript 模拟操作系统的最短作业优先技术,给定作业/进程、到达时间和突发时间:
检查此链接:Shortest Job First - Concept .现在我有一个数组:
var arr = [
{
"job": "j4",
"at": 0,
"bt": 8
},
{
"job": "j2",
"at": 2,
"bt": 4
},
{
"job": "j3",
"at": 2,
"bt": 5
},
{
"job": "j5",
"at": 6,
"bt": 4
},
{
"job": "j1",
"at": 8,
"bt": 3
}
];
我想创建一个包含进程对象的新数组。 .
例如
[
{"job": "j4", "range" : "0-2"},
{"job": "j2", "range" : "2-6"},
{"job": "j5", "range" : "6-10"},
{"job": "j1", "range" : "10-13"},
{"job": "j3", "range" : "13-18"},
{"job": "j4", "range" : "18-24"}
]
所以我尝试这样做,但我 super 卡住了,离我想要实现的目标还差得很远。
for(i = 0; i < arr.length; i++) {
var temp = [];
for(j = arr[i].at; j < arr[i].bt; j++) {
var clone = arr.slice(0);
var arrived = clone.splice(0, i).filter(function( obj ) {
return obj.at == j;
});
var shorter = arr[i];
for(k = 0; k < arrived.length; k++) {
if(arrived[k].bt < arr[i].bt) {
shorter = arrived[k];
arr[i].bt - (j - arr[i].at);
}
}
if(shorter != arr[i]) {
j = arr[i].bt;
}
}
}
编辑 如果解决了,用实际值替换新的数组值
Time Process
0 j4(8)
1
2 j4(6), j2`(4), j3(5)
...
6 j4(6), j3(5), j5(4) ::: j2 done
7
8 j5`(2), j4(6), j3(5), j1(3)
9
10 ::: j5 done
...
13 ::: j1 done
...
18 ::: j3 done
...
24 ::: j4 done
so the new array will be
[
{"job": "j4", "range" : "0-2"},
{"job": "j2", "range" : "2-6"},
{"job": "j5", "range" : "6-10"},
{"job": "j1", "range" : "10-13"},
{"job": "j3", "range" : "13-18"},
{"job": "j4", "range" : "18-24"}
]
最佳答案
这是我用 JS 编写的一个有效的 SJF 示例。正如标签中提到的那样,我使用 jQuery 从主数组获取数据。将事件对象与队列进行比较,根据需要将元素移入和移出队列,并将运行时间添加到最终数组。希望这个示例可以帮助您使代码正常工作。
var active = undefined,
queue = [],
final = [],
totalBurst = 0;
// Get the total burst time
$.map(arr, function(job, index) {
// Add a run time variable to
job.runTime = job.bt;
totalBurst += job.bt + job.at;
});
// This loop simulates time
for (var i = 0; i < totalBurst; i+=1) {
if (typeof active === 'object') {
active.runTime -= 1;
if (active.runTime < 1) {
final.push({ job : active.job, start : active.start, end : i});
active = undefined;
}
}
// Get array of jobs recieved at this time signature
var toProcess,
jobs = $.grep(arr, function(job, index) {
return job.at === i;
});
// Merge new jobs into queue
queue = queue.concat(jobs);
// Sort the queue
queue.sort(function(a,b) {
return a.bt < b.bt ? -1 : 1;
});
// Get the job to process next
toProcess = queue.splice(0,1)[0];
if (typeof toProcess !== 'undefined') {
// Process active job
if (typeof active === 'undefined' && typeof toProcess !== 'undefined') {
// Automatically start the first job in the queue
toProcess.start = i;
active = toProcess;
} else if( typeof toProcess !== 'undefined' && active.bt > toProcess.bt ) {
// Push active time to final array
final.push({ job : active.job, start : active.start, end : i});
// If active still has time to run add it to queue
if (active.runTime > 0) {
queue.push(active);
}
// Create new active process
toProcess.start = i;
active = toProcess;
} else if( typeof toProcess !== 'undefined') {
// Otherwise we still have an active process
// Push the toProcess back on the queue
queue.push(toProcess);
}
}
}
关于javascript - 在javascript中做最短工作优先算法,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/19386474/