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将两个列表合并为一个列表并对项目进行排序

sorting c#

Manish Basantani · 技术社区 · 16 年前

有没有一种方法可以将两个给定的列表合并到一个列表中,并使用一个for循环以排序的方式存储这些项?

另外,我正在寻找一种不使用API方法的解决方案(如联合、排序等)。

示例代码。

private static void MergeAndOrder() 
{
var listOne = new List<int> {3, 4, 1, 2, 7, 6, 9, 11}; 
var listTwo = new List<int> {1, 7, 8, 3, 5, 10, 15, 12}; 

//Without Using C# helper methods...
//ToDo.............................

//Using C# APi.
var expectedResult = listOne.Union(listTwo).ToList(); 
expectedResult.Sort();//Output: 1,2,3,4,5,6,7,8,9,10,11,12,15
//I need the same result without using API methods, and that too by iterating over items only once.


}

附言:我在一次采访中被问到这个问题,但至今还没有找到答案。

8 回复 | 直到 8 年前

Jonathon Faust 16 年前

List a = { 1, 3, 5, 7, 9 }
List b = { 2, 4, 6, 8, 10 }
List result = { }
int i=0, j=0, lastIndex=0
while(i < a.length || j < b.length)
    // If we're done with a, just gobble up b (but don't add duplicates)
    if(i >= a.length)
        if(result[lastIndex] != b[j])
            result[++lastIndex] = b[j]
        j++
        continue

    // If we're done with b, just gobble up a (but don't add duplicates)
    if(j >= b.length)
        if(result[lastIndex] != a[i])
            result[++lastIndex] = a[i]
        i++
        continue

    int smallestVal

    // Choose the smaller of a or b
    if(a[i] < b[j])
        smallestVal = a[i++]
    else
        smallestVal = b[j++]

    // Don't insert duplicates
    if(result[lastIndex] != smallestVal)
        result[++lastIndex] = smallestVal
end while

Joel Coehoorn 16 年前

.ToList() .ToArray()

var expectedResult = listOne.Union(listTwo).OrderBy(i => i);

Sailaja V 8 年前

    private static void MergeTwoSortedArray(int[] first, int[] second)
    {
        //throw new NotImplementedException();

        int[] result = new int[first.Length + second.Length];

        int i=0 , j=0 , k=0;

        while(i < first.Length && j <second.Length)
        {
            if(first[i] < second[j])
            {
                result[k++] = first[i++];
            }
            else
            {
                result[k++] = second[j++];
            }
        }

        if (i < first.Length)
        {
            for (int a = i; a < first.Length; a++)
                result[k] = first[a];
        }

        if (j < second.Length)
        {
            for (int a = j; a < second.Length; a++)
                result[k++] = second[a];
        }

        foreach (int a in result)
            Console.Write(a + " ");

        Console.WriteLine();
    }

Bartek 8 年前

        var list1 = new List<int?>() {
            1,3,5,9,11
        };

        var list2 = new List<int?>() {
            2,5,6,11,15,17,19,29
        };

        foreach (var c in SortedAndMerged(list1.GetEnumerator(), list2.GetEnumerator())) {
            Console.Write(c+" ");
        }

        Console.ReadKey();
    }

    private static IEnumerable<int> SortedAndMerged(IEnumerator<int?> e1, IEnumerator<int?> e2) {
        e2.MoveNext();
        e1.MoveNext();

        do {
            while (e1.Current < e2.Current) {
                if (e1.Current != null) yield return e1.Current.Value;
                e1.MoveNext();
            }

            if (e2.Current != null) yield return e2.Current.Value;
            e2.MoveNext();
        } while (!(e1.Current == null && e2.Current == null));
    }
}

LBushkin 16 年前

Chris R. Timmons 16 年前

var listOne = new List<int> { 3, 4, 1, 2, 7, 6, 9, 11 };
var listTwo = new List<int> { 1, 7, 8, 3, 5, 10, 15, 12 };
var result = listOne.ToList();

foreach (var n in listTwo)
{
  if (result.IndexOf(n) == -1)
    result.Add(n);
}

ewernli 16 年前

int[] values = new int[ Integer.MAX ]; // initialize with 0
int size1 = list1.size();
int size2 = list2.size();

for( int pos = 0; pos < size1 + size2 ; pos++ )
{
     int val =  pos > size1 ? list2[ pos-size1 ] : list1[ pos ] ;
     values[ val ]++;
}

values 0

paulH 12 年前

List<int> sortedList = new List<int>();
foreach (int x in listOne)
{
    sortedList<x> = x;
}
foreach (int x in listTwo)
{
    sortedList<x> = x;
}