使用LINQ对没有间隙的数字序列进行分组

时间:2011-01-13 15:45:50

标签: c# linq

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"

建议? Linq?

4 个答案:

答案 0 :(得分:42)

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)));

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;
            }
        }
    }
}

答案 1 :(得分:4)

你不需要Linq;实际上,最简单的解决方案需要知道数组中的三个位置(您的起始编号,当前编号和当前编号之后的下一个编号),而Enumerables并不适合。

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;
   }  

}

您可以重新安排此操作以减少代码嵌套,continue在循环逻辑的早期,并删除一些变量;你将获得几毫秒的执行时间。在获取String之前,你还需要在StringBuilder的末尾修剪最后两个字符(一个尾随的逗号和空格)。

答案 2 :(得分:1)

您想要实施的算法是什么?弄清楚你想要发生什么,然后看看是否可以通过LINQ翻译使其更清晰。这里有一些非LINQ可以给你一个想法。

int[] array = { 1, 2, 3, 4, 7, 8, 11, 15, 16, 17, 18};
List<string> ranges = new List<string>();

// code assumes array is not zero-length, is distinct, and is sorted.
// to do: handle scenario as appropriate if assumptions not valid

Action<int, int, List<string>> addToRanges = (first, last, list) =>
{
    if (last == first)
        list.Add(last.ToString());
    else
        list.Add(string.Format("{0}-{1}", first, last)); ;
};

int firstItem = array[0];
int lastItem = firstItem;
foreach (int item in array.Skip(1))
{
    if (item > lastItem + 1)
    {
        addToRanges(firstItem, lastItem, ranges);
        firstItem = lastItem = item;
    }
    else
    {
        lastItem = item;
    }
}

addToRanges(firstItem, lastItem, ranges);

// return ranges or ranges.ToArray()

答案 3 :(得分:1)

这是一个剪切:

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);
    }
}