按顺序检查丢失的数字

Question

我有一个List<int>，其中包含1,2,4,7,9例如。

我有10

的范围为0有没有办法来确定哪些号码在顺序都缺失了？

我认为LINQ可能会提供一个选项，但我看不到一个

在现实世界中我还可以列举包含100,000个项目因此性能是关键

Answer 1

var list = new List<int>(new[] { 1, 2, 4, 7, 9 }); 
var result = Enumerable.Range(0, 10).Except(list); 

Answer 2

把你要检查的范围转换为HashSet：

public IEnumerable<int> FindMissing(IEnumerable<int> values) 
{ 
    HashSet<int> myRange = new HashSet<int>(Enumerable.Range(0,10)); 
    myRange.ExceptWith(values); 
    return myRange; 
} 

将返回不在values中的值。

Answer 3

LINQ的Except方法将是最可读的。它是否适合你的工作将是测试的问题。

E.g.

range.Except(listOfValues); 

编辑

下面是我用我的小基准测试程序，为他人堵塞逃脱：

static void Main() 
{ 
    var a = Enumerable.Range(0, 1000000); 
    var b = new List<int>(); 

    for (int i = 0; i < 1000000; i += 10) 
    { 
     b.Add(i); 
    } 

    Stopwatch sw = new Stopwatch(); 
    sw.Start(); 
    var c = a.Except(b).ToList(); 
    sw.Stop(); 
    Console.WriteLine("Milliseconds {0}", sw.ElapsedMilliseconds); 
    sw.Reset(); 

    Console.ReadLine(); 
} 

Answer 4

创建的NUM项的数组

const int numItems = 1000; 
bool found[numItems] = new bool[numItems]; 


List<int> list; 

PopulateList(list); 

list.ForEach(i => found[i] = true); 

// now iterate found for the numbers found 
for(int count = 0; i < numItems; ++numItems){ 
    Console.WriteList("Item {0} is {1}", count, found[count] ? "there" : "not there"); 
} 

Answer 5

 List<int> selectedNumbers = new List<int>(){8, 5, 3, 12, 2}; 

     int firstNumber = selectedNumbers.OrderBy(i => i).First(); 
     int lastNumber = selectedNumbers.OrderBy(i => i).Last(); 

     List<int> allNumbers = Enumerable.Range(firstNumber, lastNumber - firstNumber + 1).ToList(); 

     List<int> missingNumbers = allNumbers.Except(selectedNumbers).ToList(); 

     foreach (int i in missingNumbers) 
     { 
      Response.Write(i); 
     } 

Answer 6

如果范围是可预测的，我建议以下解决方案：

public static void Main() 
{ 
    //set up the expected range 
    var expectedRange = Enumerable.Range(0, 10); 

    //set up the current list 
    var currentList = new List<int> {1, 2, 4, 7, 9}; 

    //get the missing items 
    var missingItems = expectedRange.Except(currentList);  

    //print the missing items 
    foreach (int missingItem in missingItems) 
    { 
     Console.WriteLine(missingItem); 
    } 

    Console.ReadLine(); 
} 

问候， y00daa

Answer 7

这不使用LINQ，但在线性时间工作。

我假设输入列表是排序的。

这需要O(list.Count)。

private static IEnumerable<int> get_miss(List<int> list,int length) 
{ 
    var miss = new List<int>(); 
    int i =0; 
    for (i = 0; i < list.Count - 1; i++) 
    { 
     foreach (var item in 
        Enumerable.Range(list[i] + 1, list[i + 1] - list[i] - 1)) 
     { 
      yield return item; 
     } 

    } 
    foreach (var item in Enumerable.Range(list[i]+1,length-list[i])) 
    { 
     yield return item; 
    } 
} 

---

这应该采取O(n)其中n是全范围长度。

static void Main() 
    { 
     List<int> identifiers = new List<int>() { 1, 2, 4, 7, 9 }; 

     Stopwatch sw = new Stopwatch(); 

     sw.Start(); 
     List<int> miss = GetMiss(identifiers,150000); 
     sw.Stop(); 
     Console.WriteLine("{0}",sw.ElapsedMilliseconds); 

    } 
private static List<int> GetMiss(List<int> identifiers,int length) 
{ 
    List<int> miss = new List<int>(); 

    int j = 0; 

    for (int i = 0; i < length; i++) 
    { 
     if (i < identifiers[j]) 
      miss.Add(i); 

     else if (i == identifiers[j]) 
      j++; 

     if (j == identifiers.Count) 
     { 
      miss.AddRange(Enumerable.Range(i + 1, length - i)); 
      break; 
     } 
    } 

    return miss; 
} 

Answer 8

好的，真的，创建一个新的列表，它与最初的列表并行运行除了它之外......

我创建使用Aggregate方法，而不是找missings完全LINQ答案：

var list = new List<int>(new[] { 1, 2, 4, 7, 9 }); // Assumes list is ordered at this point 

list.Insert(0, 0); // No error checking, just put in the lowest and highest possibles. 
list.Add(10);  // For real world processing, put in check and if not represented then add it/them. 

var missing = new List<int>(); // Hold any missing values found. 

list.Aggregate ((seed, aggr) => // Seed is the previous #, aggr is the current number. 
{ 
    var diff = (aggr - seed) -1; // A difference between them indicates missing. 

    if (diff > 0)     // Missing found...put in the missing range. 
     missing.AddRange(Enumerable.Range((aggr - diff), diff)); 

    return aggr;  
}); 

丢失的列表中有此之后，上面的代码已被执行：

3, 5, 6, 8 

Answer 9

任何两个IEnunumerable<T>其中T :IComparable一般工作的替代方法。如果两个IE枚举都是排序的，则这在O（1）存储器（即，没有创建另一个ICollection并减去等）和在O（n）时间中起作用。

使用IEnumerable<IComparable>和GetEnumerator使这个可读性稍差，但更一般。

实施

/// <summary> 
/// <para>For two sorted IEnumerable&lt;T&gt; (superset and subset),</para> 
/// <para>returns the values in superset which are not in subset.</para> 
/// </summary> 
public static IEnumerable<T> CompareSortedEnumerables<T>(IEnumerable<T> superset, IEnumerable<T> subset) 
    where T : IComparable 
{ 
    IEnumerator<T> supersetEnumerator = superset.GetEnumerator(); 
    IEnumerator<T> subsetEnumerator = subset.GetEnumerator(); 
    bool itemsRemainingInSubset = subsetEnumerator.MoveNext(); 

    // handle the case when the first item in subset is less than the first item in superset 
    T firstInSuperset = superset.First(); 
    while (itemsRemainingInSubset && supersetEnumerator.Current.CompareTo(subsetEnumerator.Current) >= 0) 
     itemsRemainingInSubset = subsetEnumerator.MoveNext(); 

    while (supersetEnumerator.MoveNext()) 
    { 
     int comparison = supersetEnumerator.Current.CompareTo(subsetEnumerator.Current); 
     if (!itemsRemainingInSubset || comparison < 0) 
     { 
      yield return supersetEnumerator.Current; 
     } 
     else if (comparison >= 0) 
     { 
      while (itemsRemainingInSubset && supersetEnumerator.Current.CompareTo(subsetEnumerator.Current) >= 0) 
       itemsRemainingInSubset = subsetEnumerator.MoveNext(); 
     } 
    } 
} 

使用

var values = Enumerable.Range(0, 11); 
var list = new List<int> { 1, 2, 4, 7, 9 }; 
var notIncluded = CompareSortedEnumerables(values, list);