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?
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
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