从给定列表中检测至少 3 个连续数字的序列

Question

我有一个数字列表，例如 21,4,7,9,12,22,17,8,2,20,23

我希望能够挑选出连续数字的序列（长度至少为 3 个项目），所以从上面的例子是 7,8,9 和 20,21,22,23。

我尝试过一些丑陋的庞大函数，但我想知道是否有一种简洁的 LINQ 风格的方法来做到这一点。

有什么建议吗？

更新：

非常感谢您的所有回复，非常感谢。我目前正在与他们一起玩，看看哪一个最适合融入我们的项目。

Answer 1

我突然意识到你应该做的第一件事就是订购清单。然后只需遍历它，记住当前序列的长度并检测它何时结束即可。老实说，我怀疑简单的 foreach 循环将是最简单的方法 - 我无法立即想到任何非常简洁的类似 LINQ 的方法。如果您确实愿意，您当然可以在迭代器块中执行此操作，但请记住，对列表进行排序意味着无论如何您都有相当的“前期”成本。所以我的解决方案看起来像这样：

var ordered = list.OrderBy(x => x);
int count = 0;
int firstItem = 0; // Irrelevant to start with
foreach (int x in ordered)
{
    // First value in the ordered list: start of a sequence
    if (count == 0)
    {
        firstItem = x;
        count = 1;
    }
    // Skip duplicate values
    else if (x == firstItem + count - 1)
    {
        // No need to do anything
    }
    // New value contributes to sequence
    else if (x == firstItem + count)
    {
        count++;
    }
    // End of one sequence, start of another
    else
    {
        if (count >= 3)
        {
            Console.WriteLine("Found sequence of length {0} starting at {1}",
                              count, firstItem);
        }
        count = 1;
        firstItem = x;
    }
}
if (count >= 3)
{
    Console.WriteLine("Found sequence of length {0} starting at {1}",
                      count, firstItem);
}

编辑：好的，我只是想到了一种更类似于 LINQ 的做事方式。我现在没有时间完全实现它，但是：

• 排序序列
• 使用类似 SelectWithPrevious（可能更好地命名为SelectConsecutive）来获取连续的元素
• 对使用包含索引的 Select 重载来获取 (index,当前、上一个）
• 过滤掉 (当前 = 上一个 + 1) 处的任何项目，以获取算作序列开始的任何位置（特殊情况索引=0）
• 使用 SelectWithPrevious 在结果上获取两个起点之间序列的长度（从前一个索引中减去一个索引）
• 过滤掉任何长度小于3的序列

我怀疑你需要在有序序列上连接int.MinValue，以保证最终项目被正确使用。

编辑：好的，我已经实现了这个。这是我能想到的最 LINQest 方法...我使用 null 值作为“哨兵”值来强制开始和结束序列 - 有关更多详细信息，请参阅注释。

总的来说，我不会推荐这个解决方案。很难让你清醒过来，尽管我相当有信心它是正确的，但我花了一段时间思考可能存在的相差一错误等。这是一次有趣的探索你可以做什么的旅程使用 LINQ...以及您可能不应该做的事情。

哦，请注意，我已将“最小长度 3”部分推送给调用者 - 当您有这样的元组序列时，单独过滤掉它会更干净，IMO。

using System;
using System.Collections.Generic;
using System.Linq;

static class Extensions
{
    public static IEnumerable<TResult> SelectConsecutive<TSource, TResult>
        (this IEnumerable<TSource> source,
         Func<TSource, TSource, TResult> selector)
    {
        using (IEnumerator<TSource> iterator = source.GetEnumerator())
        {
           if (!iterator.MoveNext())
           {
               yield break;
           }
           TSource prev = iterator.Current;
           while (iterator.MoveNext())
           {
               TSource current = iterator.Current;
               yield return selector(prev, current);
               prev = current;
           }
        }
    }
}

class Test
{
    static void Main()
    {
        var list = new List<int> {  21,4,7,9,12,22,17,8,2,20,23 };

        foreach (var sequence in FindSequences(list).Where(x => x.Item1 >= 3))
        {
            Console.WriteLine("Found sequence of length {0} starting at {1}",
                              sequence.Item1, sequence.Item2);
        }
    }

    private static readonly int?[] End = { null };

    // Each tuple in the returned sequence is (length, first element)
    public static IEnumerable<Tuple<int, int>> FindSequences
         (IEnumerable<int> input)
    {
        // Use null values at the start and end of the ordered sequence
        // so that the first pair always starts a new sequence starting
        // with the lowest actual element, and the final pair always
        // starts a new one starting with null. That "sequence at the end"
        // is used to compute the length of the *real* final element.
        return End.Concat(input.OrderBy(x => x)
                               .Select(x => (int?) x))
                  .Concat(End)
                  // Work out consecutive pairs of items
                  .SelectConsecutive((x, y) => Tuple.Create(x, y))
                  // Remove duplicates
                  .Where(z => z.Item1 != z.Item2)
                  // Keep the index so we can tell sequence length
                  .Select((z, index) => new { z, index })
                  // Find sequence starting points
                  .Where(both => both.z.Item2 != both.z.Item1 + 1)
                  .SelectConsecutive((start1, start2) => 
                       Tuple.Create(start2.index - start1.index, 
                                    start1.z.Item2.Value));
    }
}

Answer 2

Jon Skeet / Timwi 的解决方案是正确的选择。

为了好玩，这里有一个 LINQ 查询来完成这项工作（非常效率低下）：

var sequences = input.Distinct()
                     .GroupBy(num => Enumerable.Range(num, int.MaxValue - num + 1)
                                               .TakeWhile(input.Contains)
                                               .Last())  //use the last member of the consecutive sequence as the key
                     .Where(seq => seq.Count() >= 3)
                     .Select(seq => seq.OrderBy(num => num)); // not necessary unless ordering is desirable inside each sequence.

通过将输入加载到 HashSet 中，可以稍微提高查询的性能（以提高 包含），但这仍然不会产生一个接近高效的解决方案。

我知道的唯一错误是，如果序列包含大量的负数，则可能会出现算术溢出（我们无法表示Range<的count参数） /代码>）。使用自定义 static IEnumerable可以很容易地解决这个问题。 To(this int start, int end) 扩展方法。如果有人能想到任何其他简单的避免溢出的技术，请告诉我。

编辑：
这是一个稍微详细（但同样低效）的变体，没有溢出问题。

var sequences = input.GroupBy(num => input.Where(candidate => candidate >= num)
                                          .OrderBy(candidate => candidate)
                                          .TakeWhile((candidate, index) => candidate == num + index)
                                          .Last())
                     .Where(seq => seq.Count() >= 3)
                     .Select(seq => seq.OrderBy(num => num));

Answer 3

我认为我的解决方案更加优雅和简单，因此更容易验证其正确性：

/// <summary>Returns a collection containing all consecutive sequences of
/// integers in the input collection.</summary>
/// <param name="input">The collection of integers in which to find
/// consecutive sequences.</param>
/// <param name="minLength">Minimum length that a sequence should have
/// to be returned.</param>
static IEnumerable<IEnumerable<int>> ConsecutiveSequences(
    IEnumerable<int> input, int minLength = 1)
{
    var results = new List<List<int>>();
    foreach (var i in input.OrderBy(x => x))
    {
        var existing = results.FirstOrDefault(lst => lst.Last() + 1 == i);
        if (existing == null)
            results.Add(new List<int> { i });
        else
            existing.Add(i);
    }
    return minLength <= 1 ? results :
        results.Where(lst => lst.Count >= minLength);
}

相对于其他解决方案的优点：

• 它可以找到重叠的序列。
• 它可以正确地重用并记录下来。
• 我还没有发现任何错误;-)

Answer 4

以下是如何以“LINQish”方式解决问题：

int[] arr = new int[]{ 21, 4, 7, 9, 12, 22, 17, 8, 2, 20, 23 };
IOrderedEnumerable<int> sorted = arr.OrderBy(x => x);
int cnt = sorted.Count();
int[] sortedArr = sorted.ToArray();
IEnumerable<int> selected = sortedArr.Where((x, idx) =>
    idx <= cnt - 3 && sortedArr[idx + 1] == x + 1 && sortedArr[idx + 2] == x + 2);
IEnumerable<int> result = selected.SelectMany(x => new int[] { x, x + 1, x + 2 }).Distinct();

Console.WriteLine(string.Join(",", result.Select(x=>x.ToString()).ToArray()));

由于数组复制和重建，该解决方案当然不如传统的循环解决方案高效。

Answer 5

不是 100% Linq，但这是一个通用变体：

static IEnumerable<IEnumerable<TItem>> GetSequences<TItem>(
        int minSequenceLength, 
        Func<TItem, TItem, bool> areSequential, 
        IEnumerable<TItem> items)
    where TItem : IComparable<TItem>
{
    items = items
        .OrderBy(n => n)
        .Distinct().ToArray();

    var lastSelected = default(TItem);

    var sequences =
        from startItem in items
        where startItem.Equals(items.First())
            || startItem.CompareTo(lastSelected) > 0
        let sequence =
            from item in items
            where item.Equals(startItem) || areSequential(lastSelected, item)
            select (lastSelected = item)
        where sequence.Count() >= minSequenceLength
        select sequence;

    return sequences;
}

static void UsageInt()
{
    var sequences = GetSequences(
            3,
            (a, b) => a + 1 == b,
            new[] { 21, 4, 7, 9, 12, 22, 17, 8, 2, 20, 23 });

    foreach (var sequence in sequences)
        Console.WriteLine(string.Join(", ", sequence.ToArray()));
}

static void UsageChar()
{
    var list = new List<char>(
        "abcdefghijklmnopqrstuvwxyz".ToCharArray());

    var sequences = GetSequences(
            3,
            (a, b) => (list.IndexOf(a) + 1 == list.IndexOf(b)),
            "PleaseBeGentleWithMe".ToLower().ToCharArray());

    foreach (var sequence in sequences)
        Console.WriteLine(string.Join(", ", sequence.ToArray()));
}

Answer 6

这是我的尝试：

public static class SequenceDetector
{
    public static IEnumerable<IEnumerable<T>> DetectSequenceWhere<T>(this IEnumerable<T> sequence, Func<T, T, bool> inSequenceSelector)
    {
        List<T> subsequence = null;
        // We can only have a sequence with 2 or more items
        T last = sequence.FirstOrDefault();
        foreach (var item in sequence.Skip(1))
        {
            if (inSequenceSelector(last, item))
            {
                // These form part of a sequence
                if (subsequence == null)
                {
                    subsequence = new List<T>();
                    subsequence.Add(last);
                }
                subsequence.Add(item);
            }
            else if (subsequence != null)
            {
                // We have a previous seq to return
                yield return subsequence;
                subsequence = null;
            }
            last = item;
        }
        if (subsequence != null)
        {
            // Return any trailing seq
            yield return subsequence;
        }
    }
}

public class test
{
    public static void run()
    {
        var list = new List<int> { 21, 4, 7, 9, 12, 22, 17, 8, 2, 20, 23 };
        foreach (var subsequence in list
            .OrderBy(i => i)
            .Distinct()
            .DetectSequenceWhere((first, second) => first + 1 == second)
            .Where(seq => seq.Count() >= 3))
        {
            Console.WriteLine("Found subsequence {0}", 
                string.Join(", ", subsequence.Select(i => i.ToString()).ToArray()));
        }
    }
}

这将返回形成子序列的特定项目，并允许任何类型的项目和任何标准定义，只要它可以通过比较相邻项目来确定即可。

Answer 7

对数组进行排序，然后创建另一个数组，该数组是每个元素与前一个元素之间的差异

sortedArray = 8, 9, 10, 21, 22, 23, 24, 27, 30, 31, 32
diffArray   =    1,  1, 11,  1,  1,  1,  3,  3,  1,  1

Now iterate through the difference array; if the difference equlas 1, increase the count of a variable, sequenceLength, by 1. If the difference is > 1, check the sequenceLength if it is >=2 then you have a sequence of at at least 3 consecutive elements. Then reset sequenceLenght to 0 and continue your loop on the difference array.

Answer 8

这是我在 F# 中提出的一个解决方案，将其转换为 C# LINQ 查询应该相当容易，因为 Fold 几乎等同于 LINQ 聚合运算符。

#light

let nums = [21;4;7;9;12;22;17;8;2;20;23]

let scanFunc (mainSeqLength, mainCounter, lastNum:int, subSequenceCounter:int, subSequence:'a list, foundSequences:'a list list) (num:'a) =
        (mainSeqLength, mainCounter + 1,
         num,
         (if num <> lastNum + 1 then 1 else subSequenceCounter+1),
         (if num <> lastNum + 1 then [num] else subSequence@[num]),
         if subSequenceCounter >= 3 then
            if mainSeqLength = mainCounter+1
                then foundSequences @ [subSequence@[num]]
            elif num <> lastNum + 1
                then foundSequences @ [subSequence]
            else foundSequences
         else foundSequences)

let subSequences = nums |> Seq.sort |> Seq.fold scanFunc (nums |> Seq.length, 0, 0, 0, [], []) |> fun (_,_,_,_,_,results) -> results

Answer 9

Linq 并不是万能的解决方案，有时您最好使用简单的循环。这是一个解决方案，只需一点 Linq 即可对原始序列进行排序并过滤结果

void Main()
{
    var numbers = new[] { 21,4,7,9,12,22,17,8,2,20,23 };
    var sequences =
        GetSequences(numbers, (prev, curr) => curr == prev + 1);
        .Where(s => s.Count() >= 3);
    sequences.Dump();
}

public static IEnumerable<IEnumerable<T>> GetSequences<T>(
    IEnumerable<T> source,
    Func<T, T, bool> areConsecutive)
{
    bool first = true;
    T prev = default(T);
    List<T> seq = new List<T>();
    foreach (var i in source.OrderBy(i => i))
    {
        if (!first && !areConsecutive(prev, i))
        {
            yield return seq.ToArray();
            seq.Clear();
        }
        first = false;
        seq.Add(i);
        prev = i;
    }
    if (seq.Any())
        yield return seq.ToArray();
}

Answer 10

我想到了同样的事情 Jon：要表示一系列连续整数，您真正需要的只是两个微不足道的整数！所以我将从这里开始：

struct Range : IEnumerable<int>
{
    readonly int _start;
    readonly int _count;

    public Range(int start, int count)
    {
        _start = start;
        _count = count;
    }

    public int Start
    {
        get { return _start; }
    }

    public int Count
    {
        get { return _count; }
    }

    public int End
    {
        get { return _start + _count - 1; }
    }

    public IEnumerator<int> GetEnumerator()
    {
        for (int i = 0; i < _count; ++i)
        {
            yield return _start + i;
        }
    }

    // Heck, why not?
    public static Range operator +(Range x, int y)
    {
        return new Range(x.Start, x.Count + y);
    }

    // skipping the explicit IEnumerable.GetEnumerator implementation
}

从那里，您可以编写一个静态方法来返回一堆与序列的连续数字相对应的 Range 值。

static IEnumerable<Range> FindRanges(IEnumerable<int> source, int minCount)
{
    // throw exceptions on invalid arguments, maybe...

    var ordered = source.OrderBy(x => x);

    Range r = default(Range);

    foreach (int value in ordered)
    {
        // In "real" code I would've overridden the Equals method
        // and overloaded the == operator to write something like
        // if (r == Range.Empty) here... but this works well enough
        // for now, since the only time r.Count will be 0 is on the
        // first item.
        if (r.Count == 0)
        {
            r = new Range(value, 1);
            continue;
        }

        if (value == r.End)
        {
            // skip duplicates
            continue;
        }
        else if (value == r.End + 1)
        {
            // "append" consecutive values to the range
            r += 1;
        }
        else
        {
            // return what we've got so far
            if (r.Count >= minCount)
            {
                yield return r;
            }

            // start over
            r = new Range(value, 1);
        }
    }

    // return whatever we ended up with
    if (r.Count >= minCount)
    {
        yield return r;
    }
}

演示：

int[] numbers = new[] { 21, 4, 7, 9, 12, 22, 17, 8, 2, 20, 23 };

foreach (Range r in FindConsecutiveRanges(numbers, 3))
{
    // Using .NET 3.5 here, don't have the much nicer string.Join overloads.
    Console.WriteLine(string.Join(", ", r.Select(x => x.ToString()).ToArray()));
}

输出：

7, 8, 9
20, 21, 22, 23

Answer 11

这是我对这个问题的 LINQ-y 看法：

static IEnumerable<IEnumerable<int>>
    ConsecutiveSequences(this IEnumerable<int> input, int minLength = 3)
{
    int order = 0;
    var inorder = new SortedSet<int>(input);
    return from item in new[] { new { order = 0, val = inorder.First() } }
               .Concat(
                 inorder.Zip(inorder.Skip(1), (x, val) =>
                         new { order = x + 1 == val ? order : ++order, val }))
           group item.val by item.order into list
           where list.Count() >= minLength
           select list;
}

• 不使用显式循环，但仍然应该是 O(n lg n)
• 使用 SortedSet 而不是 .OrderBy().Distinct()
• 将连续元素与 list.Zip(list.Skip(1)) 结合起来

Answer 12

这是使用字典而不是排序的解决方案......
它将项目添加到字典中，然后对于每个值在上方和下方递增以找到最长的序列。
它不是严格意义上的 LINQ，尽管它确实使用了一些 LINQ 函数，而且我认为它比纯 LINQ 解决方案更具可读性。

static void Main(string[] args)
    {
        var items = new[] { -1, 0, 1, 21, -2, 4, 7, 9, 12, 22, 17, 8, 2, 20, 23 };
        IEnumerable<IEnumerable<int>> sequences = FindSequences(items, 3);

        foreach (var sequence in sequences)
        {   //print results to consol
            Console.Out.WriteLine(sequence.Select(num => num.ToString()).Aggregate((a, b) => a + "," + b));
        }
        Console.ReadLine();
    }

    private static IEnumerable<IEnumerable<int>> FindSequences(IEnumerable<int> items, int minSequenceLength)
    {
        //Convert item list to dictionary
        var itemDict = new Dictionary<int, int>();
        foreach (int val in items)
        {
            itemDict[val] = val;
        }
        var allSequences = new List<List<int>>();
        //for each val in items, find longest sequence including that value
        foreach (var item in items)
        {
            var sequence = FindLongestSequenceIncludingValue(itemDict, item);
            allSequences.Add(sequence);
            //remove items from dict to prevent duplicate sequences
            sequence.ForEach(i => itemDict.Remove(i));
        }
        //return only sequences longer than 3
        return allSequences.Where(sequence => sequence.Count >= minSequenceLength).ToList();
    }

    //Find sequence around start param value
    private static List<int> FindLongestSequenceIncludingValue(Dictionary<int, int> itemDict, int value)
    {
        var result = new List<int>();
        //check if num exists in dictionary
        if (!itemDict.ContainsKey(value))
            return result;

        //initialize sequence list
        result.Add(value);

        //find values greater than starting value
        //and add to end of sequence
        var indexUp = value + 1;
        while (itemDict.ContainsKey(indexUp))
        {
            result.Add(itemDict[indexUp]);
            indexUp++;
        }

        //find values lower than starting value 
        //and add to start of sequence
        var indexDown = value - 1;
        while (itemDict.ContainsKey(indexDown))
        {
            result.Insert(0, itemDict[indexDown]);
            indexDown--;
        }
        return result;
    }