计算两个重复列表的差异 [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));

我想从 list1 中删除包含在 list2 中的元素，次数与它们包含在 list2 中的次数相同.在上面的例子中: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](只有一次出现 2 应该是从 list1 中删除，因为 list2 只包含 2 的一个实例).

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).

我尝试使用 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?同时遍历两个列表对我来说有点难看.

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，您只想从 list1 中删除单个 2 元素而不是全部其中.您可以遍历 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，还有比这更有效的方法.

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 的答案以获取相同的解决方案，但使用 lambda 而不是方法引用.

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

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