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 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

其他回答

如果字符串只包含ASCII字符,则可以使用此方法。

    public static string ASCIIReverse(string s)
    {
        byte[] reversed = new byte[s.Length];

        int k = 0;
        for (int i = s.Length - 1; i >= 0; i--)
        {
            reversed[k++] = (byte)s[i];
        }

        return Encoding.ASCII.GetString(reversed);
    }

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);
    }

如果在面试中你被告知不能使用Array。反过来,我想这可能是最快的一个。它不会创建新的字符串,只迭代数组的一半以上(即O(n/2)次迭代)

    public static string ReverseString(string stringToReverse)
    {
        char[] charArray = stringToReverse.ToCharArray();
        int len = charArray.Length-1;
        int mid = len / 2;

        for (int i = 0; i < mid; i++)
        {
            char tmp = charArray[i];
            charArray[i] = charArray[len - i];
            charArray[len - i] = tmp;
        }
        return new string(charArray);
    }

我们可以使用两个指针,一个指向字符串的开始,另一个指向字符串的结束。然后每次交换第i个和第j个指针的值,第i个指针的值递增+1,第j个指针的值递减-1。

string s = Console.ReadLine();
Console.WriteLine(s + "\n");
char[] charArray = s.ToCharArray();
int i = 0, j = s.Length - 1;
while (i < j) {
    char temp = charArray[i];
    charArray[i] = charArray[j];
    charArray[j] = temp;
    i++; j--;
}
string ans = new string(charArray);
Console.WriteLine(ans + "\n");
// Input: hello
// Output: olleh

使用Substring怎么样

static string ReverseString(string text)
{
    string sub = "";
    int indexCount = text.Length - 1;
    for (int i = indexCount; i > -1; i--)
    {
        sub = sub + text.Substring(i, 1);
    }
    return sub;
}