如何删除数组中的项目并保持顺序不变？

使用该行，您基本上可以获取数组中的最后一项，并将其放置在已删除的项的位置。bag[i] = bag[--capacity];

假设您使用的是（而不是 ），我们不能将数组的索引指定为 。我们能做的最好的事情就是分配它或创建一个新数组。int[]Integer[]null-1

我决定从头开始创建一个新数组。以下方法就可以了。

public class IntBag2 {
    private static final int INITIAL_SIZE = 20;
    private static int[] bag;
    private int capacity;

    public IntBag2() {
        bag = new int[INITIAL_SIZE];
    }

    public IntBag2(int capacity) {
        bag = new int[capacity];
    }

    public boolean add(int item) {
        if (capacity == bag.length)
            return false;

        bag[capacity++] = item;

        return true;
    }

    public boolean delete(int item) {
        int[] newBag = new int[capacity];
        int newCapacity = capacity;
        boolean deleted = false;
        for (int i = 0, j = 0; i < capacity; i++) {
            if (bag[i] == item && !deleted) {
                deleted = true;
                newCapacity = capacity - 1;
            } else {
                newBag[j++] = bag[i];
            }
        }

        bag = newBag;
        capacity = newCapacity;
        return deleted;
    }

    @Override
    public String toString() {
        String result = "Bag: ";
        for (int i = 0; i < capacity; i++)
            result += bag[i] + " ";
        return result;
    }
}

一个更简单的实现是可能的，允许“删除”数组中与给定条目相等的所有条目，并有效地将剩余的条目移到前面：deleteitem

public boolean delete(int item) {
    System.out.println("deleting " + item); // for debug purposes
    int oldCapacity = capacity;
    for (int i = 0, j = 0; i < oldCapacity; i++) {
        if (bag[i] != item) {
            bag[j++] = bag[i];
        } else {
            capacity--;
        }
    }
    System.out.println("new capacity = " + capacity); // for debug

    // or use Arrays.fill(bag, capacity, oldCapacity, -1); instead of the loop
    for (int i = capacity; i < oldCapacity; i++) {
        bag[i] = -1; // mark free entries with -1 in the tail
    }

    return oldCapacity == capacity;
}

此外，如果在循环中使用多个串联，则应使用方法，或者为了简洁起见，可以使用以下实用程序方法实现：toStringStringBuilderprintArrays

public void print() {
    System.out.println(Arrays.toString(Arrays.copyOf(bag, capacity)));
}

测试：

IntBag ibag = new IntBag(10);
ibag.add(1);
ibag.add(3);
ibag.add(3);
ibag.add(2);
ibag.add(1);
ibag.print();
ibag.delete(2);
ibag.print();
ibag.delete(1);
ibag.print();

输出：

[1, 3, 3, 2, 1]
deleting 2
new capacity = 4
[1, 3, 3, 1]
deleting 1
new capacity = 2
[3, 3]

更新

仅“删除”第一个条目而不创建新数组，可以按如下方式实现：

• 跳过所有元素，直到检测到或达到结束itembag
• 如果找到，将剩余元素移位 1，将 -1 写入最后一个元素，返回itemtrue
• 否则返回false

public boolean deleteFirst(int item) {
    System.out.println("deleting first " + item);
    int id = 0;
    while (id < capacity && bag[id] != item) id++;
    if (id < capacity && bag[id] == item) {
        while (++id < capacity) {
            bag[id - 1] = bag[id];
        }
        bag[--capacity] = -1;
        return true;
    }
    return false;
}

测试：

IntBag ibag = new IntBag(10);
ibag.add(1); ibag.add(3); ibag.add(1); ibag.add(2); ibag.add(1);
ibag.print();
ibag.deleteFirst(1); ibag.print();
ibag.deleteFirst(1); ibag.print();
ibag.deleteFirst(1); ibag.print();

输出：

[1, 3, 1, 2, 1]
deleting first 1
[3, 1, 2, 1]
deleting first 1
[3, 2, 1]
deleting first 1
[3, 2]