我刚刚在c# 2.0中写了一个字符串反向函数(即LINQ不可用),然后想到了这个:
public string Reverse(string text)
{
char[] cArray = text.ToCharArray();
string reverse = String.Empty;
for (int i = cArray.Length - 1; i > -1; i--)
{
reverse += cArray[i];
}
return reverse;
}
就我个人而言,我并不喜欢这个功能,我相信有更好的方法来实现它。是吗?
抱歉写了这么长时间,但这可能会很有趣
using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Text;
namespace ConsoleApplication1
{
class Program
{
public static string ReverseUsingArrayClass(string text)
{
char[] chars = text.ToCharArray();
Array.Reverse(chars);
return new string(chars);
}
public static string ReverseUsingCharacterBuffer(string text)
{
char[] charArray = new char[text.Length];
int inputStrLength = text.Length - 1;
for (int idx = 0; idx <= inputStrLength; idx++)
{
charArray[idx] = text[inputStrLength - idx];
}
return new string(charArray);
}
public static string ReverseUsingStringBuilder(string text)
{
if (string.IsNullOrEmpty(text))
{
return text;
}
StringBuilder builder = new StringBuilder(text.Length);
for (int i = text.Length - 1; i >= 0; i--)
{
builder.Append(text[i]);
}
return builder.ToString();
}
private static string ReverseUsingStack(string input)
{
Stack<char> resultStack = new Stack<char>();
foreach (char c in input)
{
resultStack.Push(c);
}
StringBuilder sb = new StringBuilder();
while (resultStack.Count > 0)
{
sb.Append(resultStack.Pop());
}
return sb.ToString();
}
public static string ReverseUsingXOR(string text)
{
char[] charArray = text.ToCharArray();
int length = text.Length - 1;
for (int i = 0; i < length; i++, length--)
{
charArray[i] ^= charArray[length];
charArray[length] ^= charArray[i];
charArray[i] ^= charArray[length];
}
return new string(charArray);
}
static void Main(string[] args)
{
string testString = string.Join(";", new string[] {
new string('a', 100),
new string('b', 101),
new string('c', 102),
new string('d', 103),
});
int cycleCount = 100000;
Stopwatch stopwatch = new Stopwatch();
stopwatch.Start();
for (int i = 0; i < cycleCount; i++)
{
ReverseUsingCharacterBuffer(testString);
}
stopwatch.Stop();
Console.WriteLine("ReverseUsingCharacterBuffer: " + stopwatch.ElapsedMilliseconds + "ms");
stopwatch.Reset();
stopwatch.Start();
for (int i = 0; i < cycleCount; i++)
{
ReverseUsingArrayClass(testString);
}
stopwatch.Stop();
Console.WriteLine("ReverseUsingArrayClass: " + stopwatch.ElapsedMilliseconds + "ms");
stopwatch.Reset();
stopwatch.Start();
for (int i = 0; i < cycleCount; i++)
{
ReverseUsingStringBuilder(testString);
}
stopwatch.Stop();
Console.WriteLine("ReverseUsingStringBuilder: " + stopwatch.ElapsedMilliseconds + "ms");
stopwatch.Reset();
stopwatch.Start();
for (int i = 0; i < cycleCount; i++)
{
ReverseUsingStack(testString);
}
stopwatch.Stop();
Console.WriteLine("ReverseUsingStack: " + stopwatch.ElapsedMilliseconds + "ms");
stopwatch.Reset();
stopwatch.Start();
for (int i = 0; i < cycleCount; i++)
{
ReverseUsingXOR(testString);
}
stopwatch.Stop();
Console.WriteLine("ReverseUsingXOR: " + stopwatch.ElapsedMilliseconds + "ms");
}
}
}
结果:
ReverseUsingCharacterBuffer: 346毫秒
ReverseUsingArrayClass: 87毫秒
ReverseUsingStringBuilder: 824毫秒
ReverseUsingStack: 2086毫秒
ReverseUsingXOR: 319毫秒
很抱歉在这个旧帖子上发帖。我在为面试练习一些代码。
这是我为c#设计的。在重构之前,我的第一个版本非常糟糕。
static String Reverse2(string str)
{
int strLen = str.Length, elem = strLen - 1;
char[] charA = new char[strLen];
for (int i = 0; i < strLen; i++)
{
charA[elem] = str[i];
elem--;
}
return new String(charA);
}
对比数组。相反的方法下面,它出现更快的12个字符或更少的字符串。13个字符后,数组。倒车开始变得更快,最终在速度上占主导地位。我只是想指出速度开始变化的大致位置。
static String Reverse(string str)
{
char[] charA = str.ToCharArray();
Array.Reverse(charA);
return new String(charA);
}
在字符串中有100个字符时,它比我的版本x 4快。然而,如果我知道字符串总是小于13个字符,我就会使用我所创建的字符串。
测试使用Stopwatch进行,并进行了500万次迭代。此外,我不确定我的版本是否处理代理或组合字符情况与Unicode编码。
因为我喜欢两个答案-一个是使用字符串。创建,因此高性能和低分配和另一个正确性-使用StringInfo类,我决定需要一种组合方法。这是最终的字符串反转方法:)
private static string ReverseString(string str)
{
return string.Create(str.Length, str, (chars, state) =>
{
var enumerator = StringInfo.GetTextElementEnumerator(state);
var position = state.Length;
while (enumerator.MoveNext())
{
var cluster = ((string)enumerator.Current).AsSpan();
cluster.CopyTo(chars.Slice(position - cluster.Length));
position -= cluster.Length;
}
});
}
还有一种更好的方法,使用StringInfo类的方法,它通过只返回索引来跳过Enumerator的大量字符串分配。
private static string ReverseString(string str)
{
return string.Create(str.Length, str, (chars, state) =>
{
var position = 0;
var indexes = StringInfo.ParseCombiningCharacters(state); // skips string creation
var stateSpan = state.AsSpan();
for (int len = indexes.Length, i = len - 1; i >= 0; i--)
{
var index = indexes[i];
var spanLength = i == len - 1 ? state.Length - index : indexes[i + 1] - index;
stateSpan.Slice(index, spanLength).CopyTo(chars.Slice(position));
position += spanLength;
}
});
}
与LINQ解决方案相比的一些基准测试:
String length 20:
LINQ Mean: 2,355.5 ns Allocated: 1440 B
string.Create Mean: 851.0 ns Allocated: 720 B
string.Create with indexes Mean: 466.4 ns Allocated: 168 B
String length 450:
LINQ Mean: 34.33 us Allocated: 22.98 KB
string.Create Mean: 19.13 us Allocated: 14.98 KB
string.Create with indexes Mean: 10.32 us Allocated: 2.69 KB