我有以下数字序列：
你可以看到这些数字很多。 我想缩短这个字符串。 假设该字符串包含超过20个数字，则应显示18个数字，然后是“...”和序列的最后两个数字。
我可能可以通过将这些数字添加到List<int>或HashSet<int>（在这种情况下，HashSet可能更快），但我认为这样会很慢。
StringBuilder temp = new StringBuilder();
for (...)
{
temp.Append($"{number} ");
}
var sequence = temp.ToString();
我想要的例子：
7 9 12 16 18 21 25 27 30 34 36 39 43 45 48 52 54 57 ... 952 954
如何从这个序列（字符串）中创建一个列表？
var myList = myOriginalSequence.Split(' ').ToList();
var first18Numbers = myList.Take(18);
var last2Numbers = myList.Skip(myList.Count() - 2);
if(myList.Count() > 20)
如何从列表创建新的序列字符串？
var myNewSequence = String.Join(" ", myList);
var myList = myOriginalSequence.Split(' ').ToList();
string myNewSequence;
if(myList.Count() > 20)
{
var first18Numbers = myList.Take(18);
var first18NumbersString = String.Join(" ", first18Numbers);
var last2Numbers = myList.Skip(myList.Count() - 2);
var last2NumbersString = String.Join(" ", last2Numbers);
myNewSequence = $"{first18NumbersString} ... {last2NumbersString}"
}
else
{
myNewSequence = myOriginalSequence;
}
Console.WriteLine(myNewSequence);
string.Length 不会遍历字符串的内容。 - yaakov这个版本比其他答案快大约8倍，且只分配了大约6%的内存。我认为你很难找到一个更快的版本：
static string Truncated(string input)
{
var indexOfEighteenthSpace = IndexOfCharSeekFromStart(input, ' ', 18);
if (indexOfEighteenthSpace <= 0) return input;
var indexOfSecondLastSpace = IndexOfCharSeekFromEnd(input, ' ', 2);
if (indexOfSecondLastSpace <= 0) return input;
if (indexOfSecondLastSpace <= indexOfEighteenthSpace) return input;
var leadingSegment = input.AsSpan().Slice(0, indexOfEighteenthSpace);
var trailingSegment = input.AsSpan().Slice(indexOfSecondLastSpace + 1);
return string.Concat(leadingSegment, " ... ", trailingSegment);
static int IndexOfCharSeekFromStart(string input, char value, int count)
{
var startIndex = 0;
for (var i = 0; i < count; i++)
{
startIndex = input.IndexOf(value, startIndex + 1);
if (startIndex <= 0) return startIndex;
}
return startIndex;
}
static int IndexOfCharSeekFromEnd(string input, char value, int count)
{
var endIndex = input.Length - 1;
for (var i = 0; i < count; i++)
{
endIndex = input.LastIndexOf(value, endIndex - 1);
if (endIndex <= 0) return endIndex;
}
return endIndex;
}
}
Span 表示一块连续的内存块。对字符串执行 AsSpan() 方法返回一个 ReadOnlySpan<char>，它映射到原始字符串中的底层字符。对 ReadOnlySpan 进行切片能够有效地构建子字符串，而无需分配新的字符串以容纳子字符串，相反它只是指向最初字符串所使用的原始内存。 - yaakov有一种替代方法可以避免迭代整个数字字符串，而且速度相当快。
.NET中的字符串基本上是字符数组，并且可以使用数组引用（[1..n]）单独引用。这可以通过从开头和结尾分别测试正确数量的空格来利用。
代码中没有什么好看的地方，但稍后可以进行优化（例如，确保字符串中实际上有内容，字符串已修剪等）。
如果你感到精力充沛，下面的函数也可以优化为单个函数。
string finalNumbers = GetStartNumbers(myListOfNumbers, 18);
if(finalNumbers.EndsWith(" ... "))
finalNumbers += GetEndNumbers(myListOfNumbers, 2);
public string GetStartNumbers(string listOfNumbers, int collectThisManyNumbers)
{
int spaceCounter = 0; // The current count of spaces
int charPointer = 0; // The current character in the string
// Loop through the list of numbers until we either run out of characters
// or get to the appropriate 'space' position...
while(spaceCounter < collectThisManyNumbers && charPointer <= listOfNumbers.Length)
{
// The following line will add 1 to spaceCounter if the character at the
// charPointer position is a space. The charPointer is then incremented...
spaceCounter += ( listOfNumbers[charPointer++]==' ' ? 1 : 0 );
}
// Now return our value based on the last value of charPointer. Note that
// if our string doesn't have the right number of elements, then it will
// not be suffixed with ' ... '
if(spaceCounter < collectThisManyNumbers)
return listOfNumbers.Substring(0, charPointer - 1);
else
return listOfNumbers.Substring(0, charPointer - 1) + " ... ";
}
public string GetEndNumbers(string listOfNumbers, int collectThisManyNumbers)
{
int spaceCounter = 0; // The current count of spaces
int charPointer = listOfNumbers.Length; // The current character in the string
// Loop through the list of numbers until we either run out of characters
// or get to the appropriate 'space' position...
while(spaceCounter < collectThisManyNumbers && charPointer >= 0)
{
// The following line will add 1 to spaceCounter if the character at the
// charPointer position is a space. The charPointer is then decremented...
spaceCounter += ( listOfNumbers[charPointer--]==' ' ? 1 : 0 );
}
// Now return our value based on the last value of charPointer...
return listOfNumbers.Substring(charPointer);
}
有些人认为使用++和--是不可取的，但这取决于你。如果你想做数学和逻辑方面的事情，就全力以赴吧！
请注意，此代码非常长，因为它被注释到了远处。
我认为这可能会有所帮助：
public IEnumerable<string> ShortenList(string input)
{
List<int> list = input.Split(" ").Select(x=>int.Parse(x)).ToList();
if (list.Count > 20)
{
List<string> trimmedStringList = list.Take(18).Select(x=>x.ToString()).ToList();
trimmedStringList.Add("...");
trimmedStringList.Add(list[list.Count-2].ToString());
trimmedStringList.Add(list[list.Count - 1].ToString());
return trimmedStringList;
}
return list.Select(x => x.ToString());
}
numberListString.SubString(0, IndexOfNth(numberListString, ' ', 19)) + "..."
（代码不准确，可能需要添加或减去索引或调整值（19））。
编辑：刚看到点号后面想要最后2个数字，您可以使用相同的技术！只需再次连接该结果即可。
注意：我使用了这种奇怪的技术，因为OP说他们想要快速的方法，我只是提供了一个潜在的基准选项:)
试试这个：
public string Shorten(string str, int startCount, int endCount)
{
//first remove any leading or trailing whitespace
str = str.Trim();
//find the first startCount numbers by using IndexOf space
//i.e. this counts the number of spaces from the start until startCount is achieved
int spaceCount = 1;
int startInd = str.IndexOf(' ');
while (spaceCount < startCount && startInd > -1)
{
startInd = str.IndexOf(' ',startInd +1);
spaceCount++;
}
//find the last endCount numbers by using LastIndexOf space
//i.e. this counts the number of spaces from the end until endCount is achieved
int lastSpaceCount = 1;
int lastInd = str.LastIndexOf(' ');
while (lastSpaceCount < endCount && lastInd > -1)
{
lastInd = str.LastIndexOf(' ', lastInd - 1);
lastSpaceCount++;
}
//if the start ind or end ind are -1 or if lastInd <= startIndjust return the str
//as its not long enough and so doesn't need shortening
if (startInd == -1 || lastInd == -1 || lastInd <= startInd) return str;
//otherwise return the required shortened string
return $"{str.Substring(0, startInd)} ... {str.Substring(lastInd + 1)}";
}
Console.WriteLine(Shorten("123 123 123 123 123 123 123 123 123 123 123",4,3));
is:
123 123 123 123 ... 123 123 123
18 ... 20 21结尾？ - CodeCaster