如何从列表中找到倒数第二个元素？

Question

我List<string>喜欢:

 List<String> lsRelation = new List<String>{"99","86","111","105"}.

现在我想找到数字111,它是倒数第二个字符串.

所以我试过了:

String strSecondLast=lsrelation.Last() - 2;

这不起作用.那么我怎样才能找到List的倒数第二个元素Last().

Answer 1

使用:

if (lsRelation.Count >= 2)
    secLast = lsRelation[lsRelation.Count - 2];

Answer 2

如果你知道这是一个IList<T>有索引器:

string secondLast = null;
if (lsRelation.Count >= 2)
    secondLast = lsRelation[lsRelation.Count - 2];

您可以创建一个扩展名,如:

public static T SecondLast<T>(this IEnumerable<T> items)
{
    if (items == null) throw new ArgumentNullException("items");
    IList<T> list = items as IList<T>;
    if (list != null)
    {
        int count = list.Count;
        if (count > 1)
        {
            return list[count - 2];
        }
        else
            throw new ArgumentException("Sequence must contain at least two elements.", "items");
    }
    else
    {
        try
        {
            return items.Reverse().Skip(1).First();
        } catch (InvalidOperationException)
        {
            throw new ArgumentException("Sequence must contain at least two elements.", "items");
        }
    }
}

然后你可以这样使用它:

string secondLast = lsRelation.SecondLast();

Answer 3

从C# 8.0开始，您可以使用Index访问相对于序列末尾的元素：

if (lsRelation.Count >= 2)
    secLast = lsRelation[^2];

有关更多信息，请参阅文档

Answer 4

这样做有很多选择.只提一个我还没有见过的人:

List<string> lsRelation = new List<String>{"99","86","111","105"};
String strSecondLast = lsRelation.Skip(lsRelation.Count() - 2).First();

Answer 5

你可以使用ElementAt(list.Count - 2):

List<String> lsRelation = new List<String> { "99", "86", "111", "105" };
Console.WriteLine(lsRelation.ElementAt(lsRelation.Count - 2)); // 111