Question

这就是我现在正在做的事情

items = new List<TableItem> ();
            items.Add (new TableItem (){ name = "Apple" });
            items.Add (new TableItem (){ name = "Pear" });
            items.Add (new TableItem (){ name = "Shoe" });
            items.Add (new TableItem (){ name = "Cake" });
            items.Add (new TableItem (){ name = "Vodka" });
            items.Add (new TableItem (){ name = "Alize" });
            items.Add (new TableItem (){ name = "Alizeee" });
            ........
            ........
            ........

这就是我现在正在做的事情，这样一个项目丢失，因为它的计数奇数，列表将从api填充，所以可以有任意数量的项目添加到列表中

            int number = items.Count;
            int number2 = number / 2;

            List<TableItem> firstList = items.GetRange(0, number2);
            List<TableItem> secondList = items.GetRange(number2, number2);


            listView1.Adapter = new HomeScreenAdapter(this, firstList);
            listView2.Adapter = new HomeScreenAdapter (this, secondList);

Answer 1

如果计数为奇数，则在第二个列表的长度上加1：

array(4) {
  [0]=>
  string(16) "Helllooooo I'mmm"
  [1]=>
  string(25) "<strong>theeeeee</strong>"
  [2]=>
  string(20) "<em> woooooorrd</em>"
  [3]=>
  string(22) "theeee loooonnngessttt"
}

Answer 2

如果您期望奇数列表您应该使用Math.Ceil或Math.Floor根据您的要求选择中间数字所以，如果你的清单包含5个元素，那么

number = Math.Ceil(5/2) // will return 3;
number = Math.Floor(5/2) // will return 2

Answer 3

您可以使用LINQ：

var half = items.Count() / 2;
var firstList = items.Take(half).ToList();
var secondList = items.Skip(half).ToList()

如何均匀拆分清单/

3 个答案: