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
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);
}
其他回答
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);
}
看看维基百科的条目。它们实现了字符串。反向扩展法。这允许你编写这样的代码:
string s = "olleh";
s.Reverse();
他们还使用ToCharArray/Reverse组合，这是这个问题的其他答案所建议的。源代码如下所示:
public static string Reverse(this string input)
{
char[] chars = input.ToCharArray();
Array.Reverse(chars);
return new String(chars);
}
string original = "Stack Overflow";
string reversed = new string(original.Reverse().ToArray());
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
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
