排除一个特定值时list.sort（）如何工作？

Question

I have a list

List<String> names = new ArrayList<String>();
names.add("One");
names.add("Two");
names.add("Three");
names.add("Six");
names.add("Four");
names.add("Five");

I want this list to sort such that Six comes first, and then it should be in alphabetical order.

names.sort((x, y) -> {
            if (x.startsWith("S"))
                return -1;
            if (y.startsWith("S"))
                return 1; 
            return x.compareTo(y);
        });

This works fine. Why it is not working with only x.startswith("S") condition. Why I need to add condition for y.startsWith("S")

Answer 1

You don't know in which order the sort algorithm will compare your String s. It might call compare("Two","Six") and it might call compare("Six","Two") . Therefore your Comparator must support both cases.

Besides, without checking startsWith("S") for both the first and second arguments, you are violating the Comparator contract, which requires that sgn(compare(x, y)) == -sgn(compare(y, x)) .

BTW, it would be better to check equals("Six") instead of startsWith("S") . Otherwise your code will fail if your input will contain "Seven".

Answer 2

collections.sort使用合并排序，合并排序的思想是通过阶段将列表划分为单个元素列表，并在这些阶段中按有序顺序将它们合并，因此不能保证您的“六个”将是第一个元素或第二个元素在比较中