我不确定为什么要使用barList中元素的索引来查找地图positions.

I'm not sure why you are using the index of an element in barList to look into the map positions.

这应该对您有帮助

Collections.sort(barList, new Comparator<Bar>() {
    @Override
    public int compare(Bar o1, Bar o2) {
        return positions.get(o1.getId()) - positions.get(o2.getId());
    }
});

可以通过单线简化此操作

This can be simplified with a one-liner

Collections.sort(barList, Comparator.comparingInt(bar -> positions.get(bar.getId())));

基本上，问题可以归结为:

Basically, the problem boils down to this:

给出两个整数列表A = {a ₁ ，a ₂ ... a _n }，B = {b ₁ ，b ₂ ，... b _m }，根据元素在第一个列表A中的出现位置对列表B进行排序

Given two lists of integers A = {a₁, a₂...a_n} and B = {b₁, b₂, ...b_m}, sort the list B based on the position of occurrence of the element in the first list, A.

对于B中的两个元素 x ， y

For two elements x, y in B

• x> y ，如果 x 在A中的 y 之前出现.
• x< y ，如果 x 在A中的 y 之后出现.
• x = y ，如果 x = y

• x > y, if x appears before y in A.
• x < y, if x appears after y in A.
• x = y, if x = y

因此，Bar的比较器函数必须比较Foo中特定元素出现的位置(基于上述内容).

So, the comparator function for Bar has to compare the position at which a particular element has appeared in Foo (based on the above).

注意::这假设(如您所说)假设Bar中没有没有元素，而Foo中没有元素. (Bar中的元素是Foo中元素的子集.)

NOTE: This assumes (as you have said) that there is no element in Bar that is not there in Foo. (The elements in Bar are a subset of the elements in Foo).