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
public static string Reverse(string input)
{
return string.Concat(Enumerable.Reverse(input));
}
当然，您可以使用Reverse方法扩展字符串类
public static class StringExtensions
{
public static string Reverse(this string input)
{
return string.Concat(Enumerable.Reverse(input));
}
}
其他回答
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。
简单而漂亮的答案是使用扩展方法:
static class ExtentionMethodCollection
{
public static string Inverse(this string @base)
{
return new string(@base.Reverse().ToArray());
}
}
这是输出:
string Answer = "12345".Inverse(); // = "54321"
处理所有类型的unicode字符
使用System.Globalization;
public static string ReverseString(this string content) {
var textElementEnumerator = StringInfo.GetTextElementEnumerator(content);
var SbBuilder = new StringBuilder(content.Length);
while (textElementEnumerator.MoveNext()) {
SbBuilder.Insert(0, textElementEnumerator.GetTextElement());
}
return SbBuilder.ToString();
}
这里有一个解决方案，正确地将字符串“Les Mise\u0301rables”反转为“selbare\u0301siM seL”。这应该像selbarésiM seL一样呈现，而不是selbaŕesiM seL(注意重音的位置)，就像大多数基于代码单元(Array。反向，等等)，甚至代码点(反向时要特别注意代理项对)。
using System;
using System.Collections.Generic;
using System.Globalization;
using System.Linq;
public static class Test
{
private static IEnumerable<string> GraphemeClusters(this string s) {
var enumerator = StringInfo.GetTextElementEnumerator(s);
while(enumerator.MoveNext()) {
yield return (string)enumerator.Current;
}
}
private static string ReverseGraphemeClusters(this string s) {
return string.Join("", s.GraphemeClusters().Reverse().ToArray());
}
public static void Main()
{
var s = "Les Mise\u0301rables";
var r = s.ReverseGraphemeClusters();
Console.WriteLine(r);
}
}
(和现场运行的例子在这里:https://ideone.com/DqAeMJ)
它只是简单地使用。net API进行字素簇迭代，它一直存在，但看起来有点“隐藏”。
必须提交一个递归的例子:
private static string Reverse(string str)
{
if (str.IsNullOrEmpty(str) || str.Length == 1)
return str;
else
return str[str.Length - 1] + Reverse(str.Substring(0, str.Length - 1));
}
