Java - outOfMemory - 堆空間

Question

因此，我正在嘗試完成一個練習，要求我實現一種在 ArrayList 對象中進行二進制搜索的方法。 從練習：

Binary search

In the Main-class, implement a method public static int binarySearch(ArrayList<Book> books, int searchedId), which searches the list it received as a parameter, for a book with an id variable that matches the value of searchedId variable it received as a parameter. If that book is found the method, should return the index it's located at, in the list it received as a parameter. If the book isn't found, the method should return the value -1.

The method must be implemented as a binary search, which assumes the list is ordered. You should also assume, that the ids towards the beginning of the list, are always smaller than the ids towards the end of the list.

我創建了兩種方法，一種是檢查 arraylist 是否已排序（ isItSorted ），另一種是在上述方法評估為真時執行二分搜索（ binarySearch ）。 請看下面：

public static boolean isItSorted(ArrayList<Book> books) {
        ArrayList<String> boo = new ArrayList<>();
        String isItSorted = "";
        for (int i = 0; i < books.size(); i++) {
            for (int j = i + 1; j < books.size(); j++) {
                if (books.get(i).getId() < books.get(j).getId()) {
                    isItSorted = "true";
                    boo.add(isItSorted);
                } else {
                    isItSorted = "false";
                    boo.add(isItSorted);
                }
            }
        }
        if (!(boo.contains("false"))) {
            return true;
        }
        return false;
    }

    public static int binarySearch(ArrayList<Book> books, long searchedId) {
        if (searchedId < 0 || books.isEmpty()) {
            return -1;
        } else if (isItSorted(books)) {
            int start = 0;
            int end = books.size() - 1;
            int middle = (start + end) / 2;

            if (books.get(middle).getId() == searchedId) {
                return middle;
            } else if (books.get(middle).getId() > searchedId) {
                end = middle - 1;
            } else if (books.get(middle).getId() < searchedId) {
                start = middle + 1;
            }
            
            while (start <= end) {
                if (books.get(start).getId() == searchedId) {
                    return start;
                }
                start++;
            }
        }
        return -1;
    }

在這些 java 文件中，有一個測試 package 可以測試我的解決方案是否正確。 雖然 95% 的測試是成功的，但當它到達下面的方法時（它將執行時間與我的其他方法（線性搜索）進行比較），我收到錯誤Java outOfMemory heap Space 。 我使用 NetBeans。 我已經嘗試過 JVM 命令。 我的解決方案似乎適用於我嘗試過的所有對象，所以下面的測試代碼可能有問題？

@Test
    @Points("07-05.2")
    public void binarySearchIsFasterThanLinearSearch() throws Throwable {
        ArrayList<Book> books = generateBooks(10000);
        Collections.sort(books, (k1, k2) -> k1.getId() - k2.getId());

        int searched = 1000001;
        long bSearchStart = System.nanoTime();
        int binarySearchId = Searching.binarySearch(books, searched);
        long bSearchEnd = System.nanoTime();
        assertTrue("When binary search does not find what was searched for, it must return -1", binarySearchId == -1);
        long lSearchStart = System.nanoTime();
        int linearSearchId = Searching.linearSearch(books, searched);
        long lSearchEnd = System.nanoTime();
        assertTrue("When linear search does not find what was searched for, it must return -1", linearSearchId == -1);

        long bSearchTime = bSearchEnd - bSearchStart;
        long lSearchTime = lSearchEnd - lSearchStart;

        assertTrue("When there are 10000 books to search, and the searched book is not found, binary search should be a lot faster than linear search. Current this isn't so", bSearchTime * 2 < lSearchTime);
    }

Answer 1

        ArrayList<String> boo = new ArrayList<>();
        String isItSorted = "";
        for (int i = 0; i < books.size(); i++) {
            for (int j = i + 1; j < books.size(); j++) {
                if (books.get(i).getId() < books.get(j).getId()) {
                    isItSorted = "true";
                    boo.add(isItSorted);
                } else {
                    isItSorted = "false";
                    boo.add(isItSorted);
                }
            }
        }

ArrayList boo 的訂單增加了 1 億件。

如果您想檢查某些內容是否已排序，您可以使用更簡單的代碼：

Book prev = books[0];
for (int i = 1; i < books.size(); i++) {
   if (prev.getId() > books[i].getId()) 
      return false;
}
return true;

但是您不需要在 binarySearch() 中調用它，因為這會破壞 binarySearch() 的目的，並使其與 linearSearch() 一樣慢。