为什么当 Comparator.compare 相等时我们需要返回 0

Question

我知道当实现Comparator接口的compare方法时我们需要返回

• 如果 o1 > o2，+1

• -1 如果 o1 < o2

• 0 如果 o1 == o2

我的问题是为什么当两者相等时我们需要返回 0？用例是什么或在哪里使用？如果我们考虑当o2大于o1或o2等于o1时排序不会改变它的位置。谁能来解释一下实际用例吗？

Java 文档说

比较其两个参数的顺序。当第一个参数小于、等于或大于第二个参数时，返回负整数、零或正整数。

这是否意味着 return -1 或 return 0 具有相同的影响？

零或正整数

 @Override

    public int compare(Test f1, Test f2) {

        if (f1.getId() > f2.getId()) {

            return 1;

        } else if (f1.getId() < f2.getId()) {

            return -1;

        } else {

            return 0;

        }

    }

Answer 1

当您排序时，-1和0本质上对排序列表的排序有非常相似的影响，因为评估compareTo为 0 的项目将被分组在一起。您“实际上”会在其他场景中使用此比较，例如您可能不想将复杂对象重复添加到列表中（是的，您也可以通过使用 a 来实现此场景）set。假设我们有一个对象Book如下：import java.util.Comparator;public class Book implements Comparable {&nbsp; String isbn;&nbsp; String title;&nbsp; public Book(String id, String title) {&nbsp; &nbsp; this.isbn = id;&nbsp; &nbsp; this.title = title;&nbsp; }&nbsp; String getIsbn() {&nbsp; &nbsp; return isbn;&nbsp; }&nbsp; String getTitle() {&nbsp; &nbsp; return title;&nbsp; }&nbsp; @Override&nbsp; public int compareTo(Object o) {&nbsp; &nbsp; return Comparator&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .comparing(Book::getIsbn)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .thenComparing(Book::getTitle)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .compare(this, (Book) o);&nbsp; }&nbsp; @Override&nbsp; public&nbsp; String toString() {&nbsp; &nbsp; String output = new StringBuilder()&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .append(isbn).append(":").append(title)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .toString();&nbsp; &nbsp; return output;&nbsp; }}在这里，我们重写了compareToof book 以创建自定义比较，首先检查书籍的 isbn，然后检查其标题。假设（例如）您有一个图书馆，里面有书籍。您可能想阻止您的用户在该图书馆中添加重复的书籍......public class Library {&nbsp; public static void main(String [] args) {&nbsp; &nbsp; List<Book> library = new ArrayList<>();&nbsp; &nbsp; library.add(new Book("9780593098240", "Children of Dune"));&nbsp; &nbsp; library.add(new Book("9780593098233", "Dune Messiah"));&nbsp; &nbsp; library.add(new Book("9780441172719", "Dune"));&nbsp; &nbsp; // Just to show the sorting, based on multiple attributes.&nbsp; &nbsp; Collections.sort(library);&nbsp; &nbsp; System.out.println("Books in library: " + Arrays.toString(library.toArray()));&nbsp; &nbsp; // You would obviously have some code for entering a book here, but easier to just create the object for an example.&nbsp;&nbsp; &nbsp; Book newBook = new Book("9780593098240", "Children of Dune");&nbsp; &nbsp; for (Book bookInLibrary : library) {&nbsp; &nbsp; &nbsp; &nbsp; if (bookInLibrary.compareTo(newBook) == 0) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; System.out.println("We already have that book in the library.");&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; break;&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }&nbsp; }}

Answer 2

如果当比较的两个值相等时返回 -1，compare(f1,f2)并且compare(f2,f1)都将返回-1。这意味着元素的顺序将不一致。它可能会破坏一些排序算法。这就是为什么总合同compare要求：sign(compare(f1,f2))&nbsp;=&nbsp;-sign(compare(f2,f1))这意味着当两个值相等时必须返回 0。

Answer 3

您可以考虑具有以下实现的二分搜索算法：function binary_search(A, n, T):    L := 0    R := n − 1    while L <= R:        m := floor((L + R) / 2)        if A[m] < T:            L := m + 1        else if A[m] > T:            R := m - 1        else:            return m    return unsuccessful假设有一个Comapator对于相等和更少的情况返回相同的值：public int compare(Test f1, Test f2) {        if (f1.getId() > f2.getId()) {            return 1;        } else {            return -1; }现在A[m] < T或A[m].compareTo(T) < 0，将是true当T等于A[m]和当A[m]小于时T。所以在这种情况下：1 2 3 4 // array and A[m] is 22 // target T2.compareTo(2)返回-1使得算法进入下一次执行L = m + 1 -> 而不是返回正确的值。事实上，二分查找会陷入不定式循环，从 2.compareTo(2)and跳转3.compareTo(2)。我希望这有帮助。