如何分割清單均勻/

Question

這是我CAM做現在

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" }); 
      ........ 
      ........ 
      ........ 

這是我在做什麼，現在，有這樣一個項目，因爲它的奇數爲計數，列表中缺少會從API填充所以有可能是添加任何數量的項目到列表

  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

加1到你的第二個列表的長度如果計數爲奇數：

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

     List<TableItem> firstList = items.GetRange(0, number2); 
     List<TableItem> secondList = items.GetRange(number2, number % 2 == 0? number2 : number2 + 1); 


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

Answer 2

如果你期望奇列表，然後 你應該使用Math.Ceil或Math.Floor根據您的需要選擇中間的數字
所以如果你的列表中包含5個元素則

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

Answer 3

你可以使用LINQ：

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