使用LINQ对没有间隙的数字序列进行分组

Question

使用此数组int[]{ 1, 2, 3, 4, 7, 8, 11, 15,16,17,18 };，如何转换为此字符串数组"1-4","7-8","11","15-18"

有什么建议吗？林克？

Answer 1

var array = new int[] { 1, 2, 3, 4, 7, 8, 11, 15, 16, 17, 18 };

var result = string.Join(",", array
    .Distinct()
    .OrderBy(x => x)
    .GroupAdjacentBy((x, y) => x + 1 == y)
    .Select(g => new int[] { g.First(), g.Last() }.Distinct())
    .Select(g => string.Join("-", g)));

使用

public static class LinqExtensions
{
    public static IEnumerable<IEnumerable<T>> GroupAdjacentBy<T>(
        this IEnumerable<T> source, Func<T, T, bool> predicate)
    {
        using (var e = source.GetEnumerator())
        {
            if (e.MoveNext())
            {
                var list = new List<T> { e.Current };
                var pred = e.Current;
                while (e.MoveNext())
                {
                    if (predicate(pred, e.Current))
                    {
                        list.Add(e.Current);
                    }
                    else
                    {
                        yield return list;
                        list = new List<T> { e.Current };
                    }
                    pred = e.Current;
                }
                yield return list;
            }
        }
    }
}

Answer 2

您不需要Linq；实际上，最简单的解决方案需要知道数组中的三个位置(起始数字、当前数字和当前数字之后的下一个数字)，而Enumerable并不适合这三个位置。

试试这个：

var start = 0;
var end = 0;
var write = false;
var builder = new StringBuilder();
for(var i=0; i<array.Length; i++)
{
   //arranged this way to avoid ArrayOutOfBoundException
   //if the next index doesn't exist or isn't one greater than the current,
   //the current index is the end of our incremental range.
   if(i+1 == array.Length || array[i+1] > array[i] + 1)
   {
      end = i;
      write = true;
   }

   if(write)
   {
      if(end - start == 0) //one number
         builder.Append(String.Format("{0}, ", array[start]);
      else //multi-number range
         builder.Append(String.Format("{0}-{1}, ", array[start], array[end]);

      start = i+1;
      end = i+1; //not really necessary but avoids any possible case of counting backwards
      write = false;
   }  

}

您可以重新安排它以减少代码嵌套，在循环逻辑的早期执行continue，并删除一些变量；您将获得几毫秒的执行时间。在取出字符串之前，还需要去掉StringBuilder末尾的最后两个字符(尾随逗号和空格)。

Answer 3

以下是它的一小段内容：

public static IEnumerable<string> ToRanges(this IEnumerable<int> values)
{
    int? start = null, end = null;
    foreach (var value in values.OrderBy(vv => vv))
    {
        if (!start.HasValue)
        {
            start = value;
        }
        else if (value == (end ?? start) + 1)
        {
            end = value;
        }
        else
        {
            yield return end.HasValue
                ? String.Format("{0}-{1}", start, end)
                : String.Format("{0}", start);
            start = value;
            end = null;
        }
    }

    if (start.HasValue)
    {
        yield return end.HasValue
            ? String.Format("{0}-{1}", start, end)
            : String.Format("{0}", start);
    }
}