计算具有重复项的两个列表的差异 [英] Calculating difference of two lists with duplicates

问题描述

我有两个列表。

List<Integer> list1 = new ArrayList<>(Arrays.asList(1, 2, 2));
List<Integer> list2 = new ArrayList<>(Arrays.asList(2, 3, 4));

我想删除 list2 来自 list1 ，正好与 list2 中包含的次数一样多。在上面的示例中：当我们删除列表1中存在于列表2中的元素时，我们应该得到结果 [1,2] （仅出现一次 2 应该从 list1 中删除​​，因为 list2 只包含一个<$的实例c $ c> 2 ）。

I want to remove the elements contained in list2 from list1, precisely as many times as they are contained in list2. In the example above: when we remove elements in list 1 which exist in list 2, we should get as result [1, 2] (only one occurrence of 2 should be removed from list1 because list2 contains only one instance of 2).

我试过 list1.removeAll（list2）; 但我得到的结果列表只包含 [1] 。

I tried with list1.removeAll(list2); but I got as result list containing only [1].

实现这一目标的最佳方法是什么？迭代两个列表同时对我来说似乎有点难看。

What is the best way to achieve this? Iterate through both lists simultaneous seems a bit ugly for me.

推荐答案

如果我理解正确，你只想删除一个 2 来自 list1 的元素，而不是所有元素。您可以迭代 list2 并尝试从 list1 中删除​​每个元素。请记住，如果 list2 不能包含重复项，则有更有效的方法。

If I understand correctly, you only want to remove a single 2 element from list1 rather than all of them. You can iterate over list2 and attempt to remove each element from list1. Keep in mind that there are more efficient methods than this if list2 cannot contain duplicates.

var list1 = new ArrayList<>(List.of(1, 2, 2));
var list2 = List.of(2, 3, 4);

list2.forEach(list1::remove);    

list1 现在包含以下内容：

[1, 2]

请参阅 starman1979的答案了解相同的解决方案，但使用的是lambda而不是方法参考。

See starman1979's answer for the same solution, but using a lambda rather than a method reference.

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