说我有这个清单：1,3,5,7,9,13
例如，给定值为：9，前一项为7，下一项为13
我如何使用C＃实现这一目标？
答案 0 :(得分：24)
您可以使用 indexer 来获取所需索引的元素。将一个添加到索引将获得 next 并从索引中减去一个将为您提供之前的元素。
int index = 4;
int prev = list[index-1];
int next = list[index+1];
您必须检查下一个和上一个索引是否存在，否则您将获得IndexOutOfRangeException异常。由于List是零基础索引，因此第一个元素将具有索引0，第二个元素将具有1，依此类推。
if(index - 1 > -1)
prev = list[index-1];
if(index + 1 < list.Length)
next = list[index+1];
答案 1 :(得分：5)
List<int> listInts = new List<int>();
listInts.AddRange(new int[] { 1, 3, 5, 7, 9, 13 });
int index = listInts.IndexOf(3); //The index here would be "1"
index++; //Check first if the index is in the length
int element = listInts[index]; //element = 5
答案 2 :(得分：5)
我通过继承.Net列表
实现了这一点public class NavigationList<T> : List<T>
{
private int _currentIndex = 0;
public int CurrentIndex
{
get
{
if (_currentIndex > Count - 1) { _currentIndex = Count - 1; }
if (_currentIndex < 0) { _currentIndex = 0; }
return _currentIndex;
}
set { _currentIndex = value; }
}
public T MoveNext
{
get { _currentIndex++; return this[CurrentIndex]; }
}
public T MovePrevious
{
get { _currentIndex--; return this[CurrentIndex]; }
}
public T Current
{
get { return this[CurrentIndex]; }
}
}
使用它变得非常容易
NavigationList<string> n = new NavigationList<string>();
n.Add("A");
n.Add("B");
n.Add("C");
n.Add("D");
Assert.AreEqual(n.Current, "A");
Assert.AreEqual(n.MoveNext, "B");
Assert.AreEqual(n.MovePrevious, "A");
答案 3 :(得分：3)
int index = list.IndexOf(9); // find the index of the given number
// find the index of next and the previous number
// by taking into account that
// the given number might be the first or the last number in the list
int prev = index > 0 ? index - 1 : -1;
int next = index < list.Count - 1 ? index + 1 : -1;
int nextItem, prevItem;
// if indexes are valid then get the items using indexer
// otherwise set them to a temporary value,
// you can also use Nullable<int> instead
nextItem = prev != -1 ? list[prev] : 0;
prevItem = next != -1 ? list[next] : 0;
答案 4 :(得分：3)
var index = list.IndexOf(9);
if (index == -1)
{
return; // or exception - whater, no element found.
}
int? nextItem = null; //null means that there is no next element.
if (index < list.Count - 1)
{
nextItem = list[index + 1];
}
int? prevItem = null;
if (index > 0)
{
prevItem = list[index - 1];
}
答案 5 :(得分：3)
在一行中使用LINQ并进行循环搜索：
下一个
rm -R node_modules/
rm yarn.lock
yarn install
上一个
YourList.SkipWhile(x => x != NextOfThisValue).Skip(1).DefaultIfEmpty( YourList[0] ).FirstOrDefault();
这是一个有效的示例(link to the fiddle)
YourList.TakeWhile(x => x != PrevOfThisValue).DefaultIfEmpty( YourList[YourList.Count-1]).LastOrDefault();
答案 6 :(得分：2)
以下可能会有所帮助
int NextValue = 0;
int PreviousValue =0;
int index = lstOfNo.FindIndex(nd =>nd.Id == 9);
var Next = lstOfNo.ElementAtOrDefault(index + 1);
var Previous = lstOfNo.ElementAtOrDefault(index - 1);
if (Next != null)
NextValue = Next;
if (Previous != null)
PreviousValue = Previous;
答案 7 :(得分：1)
这是结合 @Thunder 和 @Melad 答案获得的完整圈子列表：
private class CircularList<T> : List<T>
{
private int _currentIndex = 0;
public int CurrentIndex
{
get
{
if (_currentIndex > Count - 1) { _currentIndex = 0; }
if (_currentIndex < 0) { _currentIndex = Count - 1; }
return _currentIndex;
}
set => _currentIndex = value;
}
public int NextIndex
{
get
{
if (_currentIndex == Count - 1) return 0;
return _currentIndex + 1;
}
}
public int PreviousIndex
{
get
{
if (_currentIndex == 0) return Count - 1;
return _currentIndex - 1;
}
}
public T Next => this[NextIndex];
public T Previous => this[PreviousIndex];
public T MoveNext
{
get { _currentIndex++; return this[CurrentIndex]; }
}
public T MovePrevious
{
get { _currentIndex--; return this[CurrentIndex]; }
}
public T Current => this[CurrentIndex];
}
}
答案 8 :(得分：0)
要使其成为一种循环列表，请尝试：
public class NavigationList<T> : List<T>
{
private int _currentIndex = -1;
public int CurrentIndex
{
get
{
if (_currentIndex == Count)
_currentIndex = 0;
else if (_currentIndex > Count - 1)
_currentIndex = Count - 1;
else if (_currentIndex < 0)
_currentIndex = 0;
return _currentIndex;
}
set { _currentIndex = value; }
}
public T MoveNext
{
get { _currentIndex++; return this[CurrentIndex]; }
}
public T Current
{
get { return this[CurrentIndex]; }
}
}
答案 9 :(得分：0)
可以使用LinkedList<T>
List<int> intList = new List<int> { 1, 3, 5, 7, 9, 13 };
LinkedList<int> intLinkedList = new LinkedList<int>(intList);
Console.WriteLine("Next Value to 9 "+intLinkedList.Find(9).Next.Value);
Console.WriteLine("Next Value to 9 " +intLinkedList.Find(9).Previous.Value);
//Consider using dictionary for frequent use
var intDictionary = intLinkedList.ToDictionary(i => i, i => intLinkedList.Find(i));
Console.WriteLine("Next Value to 9 " + intDictionary[9].Next.Value);
Console.WriteLine("Next Value to 9 " + intDictionary[9].Previous.Value);
Console.Read();
答案 10 :(得分：0)
使用ElementOrDefault()
https://dotnetfiddle.net/fxVo6T
int?[] items = { 1, 3, 5, 7, 9, 13 };
for (int i = 0; i < items.Length; i++)
{
int? previous = items.ElementAtOrDefault(i - 1);
int? current = items.ElementAtOrDefault(i);
int? next = items.ElementAtOrDefault(i + 1);
}
答案 11 :(得分：0)
此外，如果您希望使用循环逻辑的紧凑型解决方案而不创建新列表，则可以使用以下代码：
int nextNumber = list[(list.IndexOf(currentNumber) + 1) % list.Count];
int previousNumber = list[(list.IndexOf(currentNumber) - 1 + list.Count) % list.Count];