在List <T>中交换两个项目

Question

是否有LINQ方式来交换两个项目的位置list<T>？

Answer 1

从C#中检查Marc的答案:Swap方法的优秀/最佳实现.

public static void Swap<T>(IList<T> list, int indexA, int indexB)
{
    T tmp = list[indexA];
    list[indexA] = list[indexB];
    list[indexB] = tmp;
}

这可以像linq-i-fied一样

public static IList<T> Swap<T>(this IList<T> list, int indexA, int indexB)
{
    T tmp = list[indexA];
    list[indexA] = list[indexB];
    list[indexB] = tmp;
    return list;
}

var lst = new List<int>() { 8, 3, 2, 4 };
lst = lst.Swap(1, 2);

Answer 2

也许有人会想到一个聪明的方法来做到这一点,但你不应该.在列表中交换两个项本身就是副作用,但LINQ操作应该是无副作用的.因此,只需使用一个简单的扩展方法:

static class IListExtensions {
    public static void Swap<T>(
        this IList<T> list,
        int firstIndex,
        int secondIndex
    ) {
        Contract.Requires(list != null);
        Contract.Requires(firstIndex >= 0 && firstIndex < list.Count);
        Contract.Requires(secondIndex >= 0 && secondIndex < list.Count);
        if (firstIndex == secondIndex) {
            return;
        }
        T temp = list[firstIndex];
        list[firstIndex] = list[secondIndex];
        list[secondIndex] = temp;
    }
}

Answer 3

从 C# 7 开始你可以做

public static IList<T> Swap<T>(IList<T> list, int indexA, int indexB)
{
    (list[indexA], list[indexB]) = (list[indexB], list[indexA]);
    return list;
}

Answer 4

没有现有的Swap方法,因此您必须自己创建一个.当然你可以使用它,但是必须考虑到一个(不成文的？)规则:LINQ操作不会改变输入参数!

在其他"linqify"答案中,(输入)列表被修改并返回,但此操作会制动该规则.如果您有一个包含未排序项的列表会很奇怪,请执行LINQ"OrderBy"操作,然后发现输入列表也是排序的(就像结果一样).这是不允许发生的!

那么......我们怎么做？

我的第一个想法是在完成迭代后恢复集合.但这是一个肮脏的解决方案,所以不要使用它:

static public IEnumerable<T> Swap1<T>(this IList<T> source, int index1, int index2)
{
    // Parameter checking is skipped in this example.

    // Swap the items.
    T temp = source[index1];
    source[index1] = source[index2];
    source[index2] = temp;

    // Return the items in the new order.
    foreach (T item in source)
        yield return item;

    // Restore the collection.
    source[index2] = source[index1];
    source[index1] = temp;
}

此解决方案很脏,因为它确实修改了输入列表,即使它将其恢复到原始状态.这可能会导致几个问题:

1. 该列表可以是只读的,这将引发异常.
2. 如果列表由多个线程共享,则在此函数期间,列表将针对其他线程进行更改.
3. 如果在迭代期间发生异常,则不会恢复列表.(这可以解决为在Swap函数中编写try-finally,并将restore-code放在finally-block中).

有一个更好(和更短)的解决方案:只需复制原始列表.(这也可以使用IEnumerable作为参数,而不是IList):

static public IEnumerable<T> Swap2<T>(this IList<T> source, int index1, int index2)
{
    // Parameter checking is skipped in this example.

    // If nothing needs to be swapped, just return the original collection.
    if (index1 == index2)
        return source;

    // Make a copy.
    List<T> copy = source.ToList();

    // Swap the items.
    T temp = copy[index1];
    copy[index1] = copy[index2];
    copy[index2] = temp;

    // Return the copy with the swapped items.
    return copy;
}

这种解决方案的一个缺点是它会复制整个列表,这将消耗内存并使解决方案相当慢.

您可以考虑以下解决方案:

static public IEnumerable<T> Swap3<T>(this IList<T> source, int index1, int index2)
{
    // Parameter checking is skipped in this example.
    // It is assumed that index1 < index2. Otherwise a check should be build in and both indexes should be swapped.

    using (IEnumerator<T> e = source.GetEnumerator())
    {
        // Iterate to the first index.
        for (int i = 0; i < index1; i++)
            yield return source[i];

        // Return the item at the second index.
        yield return source[index2];

        if (index1 != index2)
        {
            // Return the items between the first and second index.
            for (int i = index1 + 1; i < index2; i++)
                yield return source[i];

            // Return the item at the first index.
            yield return source[index1];
        }

        // Return the remaining items.
        for (int i = index2 + 1; i < source.Count; i++)
            yield return source[i];
    }
}

如果你想输入参数是IEnumerable:

static public IEnumerable<T> Swap4<T>(this IEnumerable<T> source, int index1, int index2)
{
    // Parameter checking is skipped in this example.
    // It is assumed that index1 < index2. Otherwise a check should be build in and both indexes should be swapped.

    using(IEnumerator<T> e = source.GetEnumerator())
    {
        // Iterate to the first index.
        for(int i = 0; i < index1; i++) 
        {
            if (!e.MoveNext())
                yield break;
            yield return e.Current;
        }

        if (index1 != index2)
        {
            // Remember the item at the first position.
            if (!e.MoveNext())
                yield break;
            T rememberedItem = e.Current;

            // Store the items between the first and second index in a temporary list. 
            List<T> subset = new List<T>(index2 - index1 - 1);
            for (int i = index1 + 1; i < index2; i++)
            {
                if (!e.MoveNext())
                    break;
                subset.Add(e.Current);
            }

            // Return the item at the second index.
            if (e.MoveNext())
                yield return e.Current;

            // Return the items in the subset.
            foreach (T item in subset)
                yield return item;

            // Return the first (remembered) item.
            yield return rememberedItem;
        }

        // Return the remaining items in the list.
        while (e.MoveNext())
            yield return e.Current;
    }
}

Swap4还复制了源的(子集).因此,最糟糕的情况是,它与函数Swap2一样慢且占用内存.

Answer 5

列表有反向方法.

your_list.Reverse(i, 2) // will swap elements with indexs i, i + 1. 

来源:https://msdn.microsoft.com/en-us/library/hf2ay11y(v=vs.110).aspx