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?
当前回答
这是一个非常棘手的问题。
我推荐使用Array。大多数情况下是反向的，因为它是原生编码的，维护和理解非常简单。
在我测试的所有情况下，它似乎都优于StringBuilder。
public string Reverse(string text)
{
if (text == null) return null;
// this was posted by petebob as well
char[] array = text.ToCharArray();
Array.Reverse(array);
return new String(array);
}
对于特定长度的字符串，使用Xor的第二种方法更快。
public static string ReverseXor(string s)
{
if (s == null) return null;
char[] charArray = s.ToCharArray();
int len = s.Length - 1;
for (int i = 0; i < len; i++, len--)
{
charArray[i] ^= charArray[len];
charArray[len] ^= charArray[i];
charArray[i] ^= charArray[len];
}
return new string(charArray);
}
注意:如果你想支持完整的Unicode UTF16字符集，请阅读此。而是使用那里的实现。它可以通过使用上述算法之一进一步优化，并在反转字符后遍历字符串以清理它。
下面是StringBuilder和Array之间的性能比较。逆向和异或法。
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Diagnostics;
namespace ConsoleApplication4
{
class Program
{
delegate string StringDelegate(string s);
static void Benchmark(string description, StringDelegate d, int times, string text)
{
Stopwatch sw = new Stopwatch();
sw.Start();
for (int j = 0; j < times; j++)
{
d(text);
}
sw.Stop();
Console.WriteLine("{0} Ticks {1} : called {2} times.", sw.ElapsedTicks, description, times);
}
public static string ReverseXor(string s)
{
char[] charArray = s.ToCharArray();
int len = s.Length - 1;
for (int i = 0; i < len; i++, len--)
{
charArray[i] ^= charArray[len];
charArray[len] ^= charArray[i];
charArray[i] ^= charArray[len];
}
return new string(charArray);
}
public static string ReverseSB(string text)
{
StringBuilder builder = new StringBuilder(text.Length);
for (int i = text.Length - 1; i >= 0; i--)
{
builder.Append(text[i]);
}
return builder.ToString();
}
public static string ReverseArray(string text)
{
char[] array = text.ToCharArray();
Array.Reverse(array);
return (new string(array));
}
public static string StringOfLength(int length)
{
Random random = new Random();
StringBuilder sb = new StringBuilder();
for (int i = 0; i < length; i++)
{
sb.Append(Convert.ToChar(Convert.ToInt32(Math.Floor(26 * random.NextDouble() + 65))));
}
return sb.ToString();
}
static void Main(string[] args)
{
int[] lengths = new int[] {1,10,15,25,50,75,100,1000,100000};
foreach (int l in lengths)
{
int iterations = 10000;
string text = StringOfLength(l);
Benchmark(String.Format("String Builder (Length: {0})", l), ReverseSB, iterations, text);
Benchmark(String.Format("Array.Reverse (Length: {0})", l), ReverseArray, iterations, text);
Benchmark(String.Format("Xor (Length: {0})", l), ReverseXor, iterations, text);
Console.WriteLine();
}
Console.Read();
}
}
}
以下是调查结果:
26251 Ticks String Builder (Length: 1) : called 10000 times.
33373 Ticks Array.Reverse (Length: 1) : called 10000 times.
20162 Ticks Xor (Length: 1) : called 10000 times.
51321 Ticks String Builder (Length: 10) : called 10000 times.
37105 Ticks Array.Reverse (Length: 10) : called 10000 times.
23974 Ticks Xor (Length: 10) : called 10000 times.
66570 Ticks String Builder (Length: 15) : called 10000 times.
26027 Ticks Array.Reverse (Length: 15) : called 10000 times.
24017 Ticks Xor (Length: 15) : called 10000 times.
101609 Ticks String Builder (Length: 25) : called 10000 times.
28472 Ticks Array.Reverse (Length: 25) : called 10000 times.
35355 Ticks Xor (Length: 25) : called 10000 times.
161601 Ticks String Builder (Length: 50) : called 10000 times.
35839 Ticks Array.Reverse (Length: 50) : called 10000 times.
51185 Ticks Xor (Length: 50) : called 10000 times.
230898 Ticks String Builder (Length: 75) : called 10000 times.
40628 Ticks Array.Reverse (Length: 75) : called 10000 times.
78906 Ticks Xor (Length: 75) : called 10000 times.
312017 Ticks String Builder (Length: 100) : called 10000 times.
52225 Ticks Array.Reverse (Length: 100) : called 10000 times.
110195 Ticks Xor (Length: 100) : called 10000 times.
2970691 Ticks String Builder (Length: 1000) : called 10000 times.
292094 Ticks Array.Reverse (Length: 1000) : called 10000 times.
846585 Ticks Xor (Length: 1000) : called 10000 times.
305564115 Ticks String Builder (Length: 100000) : called 10000 times.
74884495 Ticks Array.Reverse (Length: 100000) : called 10000 times.
125409674 Ticks Xor (Length: 100000) : called 10000 times.
对于短字符串，Xor似乎更快。
其他回答
这是用于反向字符串的代码
public Static void main(){
string text = "Test Text";
Console.Writeline(RevestString(text))
}
public Static string RevestString(string text){
char[] textToChar = text.ToCharArray();
string result= string.Empty;
int length = textToChar .Length;
for (int i = length; i > 0; --i)
result += textToChar[i - 1];
return result;
}
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace ConsoleApplication1
{
class Program
{
public static string ReverseString(string str)
{
int totalLength = str.Length;
int iCount = 0;
string strRev = string.Empty;
iCount = totalLength;
while (iCount != 0)
{
iCount--;
strRev += str[iCount];
}
return strRev;
}
static void Main(string[] args)
{
string str = "Punit Pandya";
string strResult = ReverseString(str);
Console.WriteLine(strResult);
Console.ReadLine();
}
}
}
首先，你不需要调用ToCharArray，因为字符串已经可以被索引为char数组，所以这将节省你的分配。
下一个优化是使用StringBuilder来防止不必要的分配(因为字符串是不可变的，连接它们每次都会复制一个字符串)。为了进一步优化这一点，我们预先设置了StringBuilder的长度，这样它就不需要扩展它的缓冲区。
public string Reverse(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();
}
编辑:性能数据
我使用Array测试了这个函数和这个函数。用下面的简单程序进行反向，其中Reverse1是一个函数，Reverse2是另一个函数:
static void Main(string[] args)
{
var text = "abcdefghijklmnopqrstuvwxyz";
// pre-jit
text = Reverse1(text);
text = Reverse2(text);
// test
var timer1 = Stopwatch.StartNew();
for (var i = 0; i < 10000000; i++)
{
text = Reverse1(text);
}
timer1.Stop();
Console.WriteLine("First: {0}", timer1.ElapsedMilliseconds);
var timer2 = Stopwatch.StartNew();
for (var i = 0; i < 10000000; i++)
{
text = Reverse2(text);
}
timer2.Stop();
Console.WriteLine("Second: {0}", timer2.ElapsedMilliseconds);
Console.ReadLine();
}
事实证明，对于短字符串数组。反向方法大约是上面方法的两倍，对于更长的字符串，差异甚至更明显。已知这个数组。反向法既简单又快，我建议你用那个而不是这个。我把这个留在这里只是为了表明这不是你应该做的方式(让我很惊讶!)
string A = null;
//a now is reversed and you can use it
A = SimulateStrReverse.StrReverse("your string");
public static class SimulateStrReverse
{
public static string StrReverse(string expression)
{
if (string.IsNullOrEmpty(expression))
return string.Empty;
string reversedString = string.Empty;
for (int charIndex = expression.Length - 1; charIndex >= 0; charIndex--)
{
reversedString += expression[charIndex];
}
return reversedString;
}
}
基于堆栈的解决方案。
public static string Reverse(string text)
{
var stack = new Stack<char>(text);
var array = new char[stack.Count];
int i = 0;
while (stack.Count != 0)
{
array[i++] = stack.Pop();
}
return new string(array);
}
Or
public static string Reverse(string text)
{
var stack = new Stack<char>(text);
return string.Join("", stack);
}
