反转字符串，甚至不使用新字符串。让说

String input  = "Mark Henry";
//Just to convert into char array. One can simply take input in char array.
Char[] array = input.toCharArray(input);
int a = input.length;

for(int i=0; i<(array.length/2 -1) ; i++)
{
    array[i] = array[i] + array[a];
    array[a] = array[i] - array[a];
    array[i] = array[i] - array[a--];
}

首先，你必须理解的是str+=将调整字符串内存大小，为1个额外的字符腾出空间。这很好，但是如果你有一本1000页的书，你想要反转，这将需要很长时间来执行。

有些人建议的解决方案是使用StringBuilder。字符串构建器在执行+=时所做的是分配更大的内存块来保存新字符，这样它就不需要在每次添加字符时进行重新分配。

如果你真的想要一个快速和最小的解决方案，我建议如下:

            char[] chars = new char[str.Length];
            for (int i = str.Length - 1, j = 0; i >= 0; --i, ++j)
            {
                chars[j] = str[i];
            }
            str = new String(chars);

在这个解决方案中，在初始化char[]时有一个初始内存分配，在string构造函数从char数组构建字符串时有一个初始内存分配。

在我的系统上，我为您运行了一个测试，反转了一个2750,000个字符的字符串。以下是10次执行的结果:

StringBuilder: 190K - 200K tick

字符数组:130K - 160K

我还运行了一个正常String +=的测试，但我在10分钟后放弃了它，没有输出。

但是，我也注意到，对于较小的字符串，StringBuilder更快，因此您必须根据输入来决定实现。

干杯

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"

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

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

就这么简单:

string x = "your string";       
string x1 = "";
for(int i = x.Length-1 ; i >= 0; i--)
    x1 += x[i];
Console.WriteLine("The reverse of the string is:\n {0}", x1);

查看输出。