Stack<T>
,Queue<T>
,List<T>
和LinkedList<T>
通过添加数据和从前端/端移除数据。以下是基准结果:
Pushing to Stack... Time used: 7067 ticks
Poping from Stack... Time used: 2508 ticks
Enqueue to Queue... Time used: 7509 ticks
Dequeue from Queue... Time used: 2973 ticks
Insert to List at the front... Time used: 5211897 ticks
RemoveAt from List at the front... Time used: 5198380 ticks
Add to List at the end... Time used: 5691 ticks
RemoveAt from List at the end... Time used: 3484 ticks
AddFirst to LinkedList... Time used: 14057 ticks
RemoveFirst from LinkedList... Time used: 5132 ticks
AddLast to LinkedList... Time used: 9294 ticks
RemoveLast from LinkedList... Time used: 4414 ticks
代码:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Diagnostics;
namespace Benchmarking
{
static class Collections
{
public static void run()
{
Random rand = new Random();
Stopwatch sw = new Stopwatch();
Stack<int> stack = new Stack<int>();
Queue<int> queue = new Queue<int>();
List<int> list1 = new List<int>();
List<int> list2 = new List<int>();
LinkedList<int> linkedlist1 = new LinkedList<int>();
LinkedList<int> linkedlist2 = new LinkedList<int>();
int dummy;
sw.Reset();
Console.Write("{0,40}", "Pushing to Stack...");
sw.Start();
for (int i = 0; i < 100000; i++)
{
stack.Push(rand.Next());
}
sw.Stop();
Console.WriteLine(" Time used: {0,9} ticks", sw.ElapsedTicks);
sw.Reset();
Console.Write("{0,40}", "Poping from Stack...");
sw.Start();
for (int i = 0; i < 100000; i++)
{
dummy = stack.Pop();
dummy++;
}
sw.Stop();
Console.WriteLine(" Time used: {0,9} ticks\n", sw.ElapsedTicks);
sw.Reset();
Console.Write("{0,40}", "Enqueue to Queue...");
sw.Start();
for (int i = 0; i < 100000; i++)
{
queue.Enqueue(rand.Next());
}
sw.Stop();
Console.WriteLine(" Time used: {0,9} ticks", sw.ElapsedTicks);
sw.Reset();
Console.Write("{0,40}", "Dequeue from Queue...");
sw.Start();
for (int i = 0; i < 100000; i++)
{
dummy = queue.Dequeue();
dummy++;
}
sw.Stop();
Console.WriteLine(" Time used: {0,9} ticks\n", sw.ElapsedTicks);
sw.Reset();
Console.Write("{0,40}", "Insert to List at the front...");
sw.Start();
for (int i = 0; i < 100000; i++)
{
list1.Insert(0, rand.Next());
}
sw.Stop();
Console.WriteLine(" Time used: {0,9} ticks", sw.ElapsedTicks);
sw.Reset();
Console.Write("{0,40}", "RemoveAt from List at the front...");
sw.Start();
for (int i = 0; i < 100000; i++)
{
dummy = list1[0];
list1.RemoveAt(0);
dummy++;
}
sw.Stop();
Console.WriteLine(" Time used: {0,9} ticks\n", sw.ElapsedTicks);
sw.Reset();
Console.Write("{0,40}", "Add to List at the end...");
sw.Start();
for (int i = 0; i < 100000; i++)
{
list2.Add(rand.Next());
}
sw.Stop();
Console.WriteLine(" Time used: {0,9} ticks", sw.ElapsedTicks);
sw.Reset();
Console.Write("{0,40}", "RemoveAt from List at the end...");
sw.Start();
for (int i = 0; i < 100000; i++)
{
dummy = list2[list2.Count - 1];
list2.RemoveAt(list2.Count - 1);
dummy++;
}
sw.Stop();
Console.WriteLine(" Time used: {0,9} ticks\n", sw.ElapsedTicks);
sw.Reset();
Console.Write("{0,40}", "AddFirst to LinkedList...");
sw.Start();
for (int i = 0; i < 100000; i++)
{
linkedlist1.AddFirst(rand.Next());
}
sw.Stop();
Console.WriteLine(" Time used: {0,9} ticks", sw.ElapsedTicks);
sw.Reset();
Console.Write("{0,40}", "RemoveFirst from LinkedList...");
sw.Start();
for (int i = 0; i < 100000; i++)
{
dummy = linkedlist1.First.Value;
linkedlist1.RemoveFirst();
dummy++;
}
sw.Stop();
Console.WriteLine(" Time used: {0,9} ticks\n", sw.ElapsedTicks);
sw.Reset();
Console.Write("{0,40}", "AddLast to LinkedList...");
sw.Start();
for (int i = 0; i < 100000; i++)
{
linkedlist2.AddLast(rand.Next());
}
sw.Stop();
Console.WriteLine(" Time used: {0,9} ticks", sw.ElapsedTicks);
sw.Reset();
Console.Write("{0,40}", "RemoveLast from LinkedList...");
sw.Start();
for (int i = 0; i < 100000; i++)
{
dummy = linkedlist2.Last.Value;
linkedlist2.RemoveLast();
dummy++;
}
sw.Stop();
Console.WriteLine(" Time used: {0,9} ticks\n", sw.ElapsedTicks);
}
}
}
如所见,Stack<T>
和Queue<T>
很快就能与之媲美,这是意料之中的。
为List<T>
,用前面和结尾有这么多的差别!令我惊讶的是,从末尾添加/删除的性能实际上与Stack<T>
.
为LinkedList<T>
,用前部操作是快速的(-比List<T>
所以有专家能解释:
Stack<T>
最后List<T>
List<T>
Last()
发布于 2018-04-02 11:58:50
L
这意味着它在内部使用“静态”数组(一个不能调整大小的数组,在.NET中称为“数组”),该数组可能而且通常大于列表的大小。然后,附加一个计数器并使用内部数组的下一个未使用的槽。如果内部数组变小以容纳所有项,则只重新分配数组(这需要复制所有元素)。当发生这种情况时,数组的大小会增加一个因子(而不是常数),通常是2。
https://stackoverflow.com/questions/-100007903
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