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;
}

其他回答

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);
    }

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));
}

使用LINQ的聚合函数

string s = "Karthik U";
s = s.Aggregate(new StringBuilder(), (o, p) => o.Insert(0, p)).ToString();

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;
}

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;
}