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
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;
}
}
其他回答
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
public string Reverse(string input)
{
char[] output = new char[input.Length];
int forwards = 0;
int backwards = input.Length - 1;
do
{
output[forwards] = input[backwards];
output[backwards] = input[forwards];
}while(++forwards <= --backwards);
return new String(output);
}
public string DotNetReverse(string input)
{
char[] toReverse = input.ToCharArray();
Array.Reverse(toReverse);
return new String(toReverse);
}
public string NaiveReverse(string input)
{
char[] outputArray = new char[input.Length];
for (int i = 0; i < input.Length; i++)
{
outputArray[i] = input[input.Length - 1 - i];
}
return new String(outputArray);
}
public string RecursiveReverse(string input)
{
return RecursiveReverseHelper(input, 0, input.Length - 1);
}
public string RecursiveReverseHelper(string input, int startIndex , int endIndex)
{
if (startIndex == endIndex)
{
return "" + input[startIndex];
}
if (endIndex - startIndex == 1)
{
return "" + input[endIndex] + input[startIndex];
}
return input[endIndex] + RecursiveReverseHelper(input, startIndex + 1, endIndex - 1) + input[startIndex];
}
void Main()
{
int[] sizes = new int[] { 10, 100, 1000, 10000 };
for(int sizeIndex = 0; sizeIndex < sizes.Length; sizeIndex++)
{
string holaMundo = "";
for(int i = 0; i < sizes[sizeIndex]; i+= 5)
{
holaMundo += "ABCDE";
}
string.Format("\n**** For size: {0} ****\n", sizes[sizeIndex]).Dump();
string odnuMaloh = DotNetReverse(holaMundo);
var stopWatch = Stopwatch.StartNew();
string result = NaiveReverse(holaMundo);
("Naive Ticks: " + stopWatch.ElapsedTicks).Dump();
stopWatch.Restart();
result = Reverse(holaMundo);
("Efficient linear Ticks: " + stopWatch.ElapsedTicks).Dump();
stopWatch.Restart();
result = RecursiveReverse(holaMundo);
("Recursive Ticks: " + stopWatch.ElapsedTicks).Dump();
stopWatch.Restart();
result = DotNetReverse(holaMundo);
("DotNet Reverse Ticks: " + stopWatch.ElapsedTicks).Dump();
}
}
输出
尺寸:10
Naive Ticks: 1
Efficient linear Ticks: 0
Recursive Ticks: 2
DotNet Reverse Ticks: 1
尺寸:100
Naive Ticks: 2
Efficient linear Ticks: 1
Recursive Ticks: 12
DotNet Reverse Ticks: 1
规格:1000
Naive Ticks: 5
Efficient linear Ticks: 2
Recursive Ticks: 358
DotNet Reverse Ticks: 9
尺寸:10000
Naive Ticks: 32
Efficient linear Ticks: 28
Recursive Ticks: 84808
DotNet Reverse Ticks: 33
尝试使用Array。反向
public string Reverse(string str)
{
char[] array = str.ToCharArray();
Array.Reverse(array);
return new string(array);
}
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));
}
如果有人问关于字符串反向的问题，其目的可能是想知道你是否知道任何位操作，比如异或。c#中有数组。反向函数，但是，您可以使用简单的异或操作在几行代码(最少)
public static string MyReverse(string s)
{
char[] charArray = s.ToCharArray();
int bgn = -1;
int end = s.Length;
while(++bgn < --end)
{
charArray[bgn] ^= charArray[end];
charArray[end] ^= charArray[bgn];
charArray[bgn] ^= charArray[end];
}
return new string(charArray);
}
