从列表中删除一个项的成本与后面要删除的项的数量成正比。在前半部分的条目符合删除条件的情况下，任何基于单独删除条目的方法最终都将不得不执行大约N*N/4个条目复制操作，如果列表很大，这可能会非常昂贵。

A faster approach is to scan through the list to find the first item to be removed (if any), and then from that point forward copy each item which should be retained to the spot where it belongs. Once this is done, if R items should be retained, the first R items in the list will be those R items, and all of the items requiring deletion will be at the end. If those items are deleted in reverse order, the system won't end up having to copy any of them, so if the list had N items of which R items, including all of the first F, were retained, it will be necessary to copy R-F items, and shrink the list by one item N-R times. All linear time.

使用for循环反向迭代列表:

for (int i = safePendingList.Count - 1; i >= 0; i--)
{
    // some code
    // safePendingList.RemoveAt(i);
}

例子:

var list = new List<int>(Enumerable.Range(1, 10));
for (int i = list.Count - 1; i >= 0; i--)
{
    if (list[i] > 5)
        list.RemoveAt(i);
}
list.ForEach(i => Console.WriteLine(i));

或者，你可以使用RemoveAll方法和一个谓词来测试:

safePendingList.RemoveAll(item => item.Value == someValue);

下面是一个简单的例子:

var list = new List<int>(Enumerable.Range(1, 10));
Console.WriteLine("Before:");
list.ForEach(i => Console.WriteLine(i));
list.RemoveAll(i => i > 5);
Console.WriteLine("After:");
list.ForEach(i => Console.WriteLine(i));

我发现自己遇到了类似的情况，我必须删除给定List<T>中的每n个元素。

for (int i = 0, j = 0, n = 3; i < list.Count; i++)
{
    if ((j + 1) % n == 0) //Check current iteration is at the nth interval
    {
        list.RemoveAt(i);
        j++; //This extra addition is necessary. Without it j will wrap
             //down to zero, which will throw off our index.
    }
    j++; //This will always advance the j counter
}

For循环是一个不好的构造。

使用时

var numbers = new List<int>(Enumerable.Range(1, 3));

while (numbers.Count > 0)
{
    numbers.RemoveAt(0);
}

但是，如果你一定要用for

var numbers = new List<int>(Enumerable.Range(1, 3));

for (; numbers.Count > 0;)
{
    numbers.RemoveAt(0);
}

或者,这个:

public static class Extensions
{

    public static IList<T> Remove<T>(
        this IList<T> numbers,
        Func<T, bool> predicate)
    {
        numbers.ForEachBackwards(predicate, (n, index) => numbers.RemoveAt(index));
        return numbers;
    }

    public static void ForEachBackwards<T>(
        this IList<T> numbers,
        Func<T, bool> predicate,
        Action<T, int> action)
    {
        for (var i = numbers.Count - 1; i >= 0; i--)
        {
            if (predicate(numbers[i]))
            {
                action(numbers[i], i);
            }
        }
    }
}

用法:

var numbers = new List<int>(Enumerable.Range(1, 10)).Remove((n) => n > 5);

然而，LINQ已经有RemoveAll()来做这件事

var numbers = new List<int>(Enumerable.Range(1, 10));
numbers.RemoveAll((n) => n > 5);

最后，你最好使用LINQ的Where()来过滤和创建一个新列表，而不是改变现有的列表。不变性通常是好的。

var numbers = new List<int>(Enumerable.Range(1, 10))
    .Where((n) => n <= 5)
    .ToList();

在遍历列表时从列表中删除项的最佳方法是使用RemoveAll()。但是人们编写的主要问题是他们必须在循环中做一些复杂的事情和/或有复杂的比较情况。

解决方案是仍然使用RemoveAll()，但使用以下符号:

var list = new List<int>(Enumerable.Range(1, 10));
list.RemoveAll(item => 
{
    // Do some complex operations here
    // Or even some operations on the items
    SomeFunction(item);
    // In the end return true if the item is to be removed. False otherwise
    return item > 5;
});

Using Remove or RemoveAt on a list while iterating over that list has intentionally been made difficult, because it is almost always the wrong thing to do. You might be able to get it working with some clever trick, but it would be extremely slow. Every time you call Remove it has to scan through the entire list to find the element you want to remove. Every time you call RemoveAt it has to move subsequent elements 1 position to the left. As such, any solution using Remove or RemoveAt, would require quadratic time, O(n²).

如果可以，使用RemoveAll。否则，下面的模式将在线性时间O(n)内就地过滤列表。

// Create a list to be filtered
IList<int> elements = new List<int>(new int[] {1, 2, 3, 4, 5, 6, 7, 8, 9, 10});
// Filter the list
int kept = 0;
for (int i = 0; i < elements.Count; i++) {
    // Test whether this is an element that we want to keep.
    if (elements[i] % 3 > 0) {
        // Add it to the list of kept elements.
        elements[kept] = elements[i];
        kept++;
    }
}
// Unfortunately IList has no Resize method. So instead we
// remove the last element of the list until: elements.Count == kept.
while (kept < elements.Count) elements.RemoveAt(elements.Count-1);