Question

我有一个看起来像这样的字符串数组

var input = new [] { "AB-PQ", "PQ-EF", "EF=CD", "CD-IJ", "IJ=XY", "XY-JK" };

我想要一个类似这样的字符串输出

var output = "AB-PQ-EF=CD-IJ=XY-JK"

我想知道是否有更好的方法，而不是使用蛮力循环，然后使用字符串生成器进行拆分和组合。

我的业务用例：我的业务用例是，给定的字符串表示包含多个城市的路线中两个城市之间的链接序列。 “-”表示道路连接，“ =”表示铁路连接。城市代码可以是任意长度。

Answer 1

这有效：

var input = new [] { "AB-PQ", "PQ-EF", "EF=CD", "CD-IJ", "IJ=XY", "XY-JK" };

var output = String.Join("",
    input.Take(1).Concat(new [] { String.Join("", input.Skip(1).Select(x => x.Substring(2))) }));

我不喜欢它，但是它可以工作。它产生"AB-PQ-EF=CD-IJ=XY-JK"。

尝试以下更可靠的方法：

void Main()
{
    var input = new[] { "AB-PQ", "PQ-XYZ", "XYZ=CD", "CD-A", "A=XY", "XY-JK" };

    var output =
        input
            .Select(i => new Segment(i))
            .Aggregate(
                "",
                (a, x) => a + x.ToString().Substring(a == "" ? 0 : x.Origin.Length));
}

public enum Mode
{
    Road, Rail
}

public sealed class Segment : IEquatable<Segment>
{
    private readonly string _origin;
    private readonly Mode _mode;
    private readonly string _destination;

    public string Origin { get { return _origin; } }
    public Mode Mode { get { return _mode; } }
    public string Destination { get { return _destination; } }

    public Segment(string descriptor)
    {
        var parts = descriptor.Split('-', '=');
        if (parts.Length != 2)
        {
            throw new System.ArgumentException("Segment descriptor must contain '=' or '-'.");
        }
        _origin = parts[0];
        _mode = descriptor.Contains("=") ? Mode.Rail : Mode.Road;
        _destination = parts[1];
    }

    public Segment(string origin, Mode mode, string destination)
    {
        _origin = origin;
        _mode = mode;
        _destination = destination;
    }

    public override bool Equals(object obj)
    {
        if (obj is Segment)
            return Equals((Segment)obj);
        return false;
    }

    public bool Equals(Segment obj)
    {
        if (obj == null) return false;
        if (!EqualityComparer<string>.Default.Equals(_origin, obj._origin)) return false;
        if (!EqualityComparer<Mode>.Default.Equals(_mode, obj._mode)) return false;
        if (!EqualityComparer<string>.Default.Equals(_destination, obj._destination)) return false;
        return true;
    }

    public override int GetHashCode()
    {
        int hash = 0;
        hash ^= EqualityComparer<string>.Default.GetHashCode(_origin);
        hash ^= EqualityComparer<Mode>.Default.GetHashCode(_mode);
        hash ^= EqualityComparer<string>.Default.GetHashCode(_destination);
        return hash;
    }

    public override string ToString()
    {
        return $"{_origin}{(_mode == Mode.Rail ? "=" : "-")}{_destination}";
    }

    public static bool operator ==(Segment left, Segment right)
    {
        if (object.ReferenceEquals(left, null))
        {
            return object.ReferenceEquals(right, null);
        }

        return left.Equals(right);
    }

    public static bool operator !=(Segment left, Segment right)
    {
        return !(left == right);
    }
}

Answer 2

您可以像这样使用LINQ和// GetJobsFromTable() runs fine and I am getting jobs to execute var dt = GetJobsFromTable(); Parallel.For(0, dt.Rows.Count, i => { var item = dt.Rows[i]; ProcessJob(item); });：

string.Join

别忘了首先将其添加到您的var input = new[]{"AB-PQ", "PQ-EF", "EF=CD", "CD-IJ", "IJ=XY", "XY-JK"}; var result = input[0] + string.Join("", input.Skip(1) .Select(c => string.Join("", c.Skip(2))));指令中：

using

Answer 3

另一个版本：

var input = new [] {"AB-PQ", "PQ-EF", "EF=CD", "CD-IJ", "IJ=XY", "XY-JK"};
var output = string.Join("=", string.Join("-", input).Split('=').Select(x => string.Join("-", x.Split('-').Distinct())))

Answer 4

var input = new [] { "AB-PQ", "PQ-EF", "EF=CD", "CD-IJ", "IJ=XY", "XY-JK" };

var r = input.Aggregate((total, current) => 
            total + current.Substring(OverlappingCharCount(total, current))
        );

您可以定义OverlappingCharCount方法以始终返回2（例如，在示例中，字符串始终重叠2个字符），或使用一些更智能的逻辑来处理更多通用情况。

Answer 5

使用Regex达到相同的目的

string str = string.Join("-", input);
Regex oRegex = new Regex(@"\b(?<Captured>\w+)-\1\b",RegexOptions.IgnoreCase);
oRegex.Matches(str).Cast<Match>().ToList().ForEach(items=>str = str.Replace(items.ToString(), items.Groups["Captured"].ToString()));
Console.WriteLine(str);

Answer 6

这是完整的答案

GUI.DrawTexture()

从字符串数组创建更大的字符串

6 个答案: