Java删除数组中的冗余项

Question

对于这个特定问题，我试图删除排序数组中的冗余元素，并在数组末尾将它们全部替换为 0 。例如，如果我有一个由 int 元素组成的数组

1,3,3,4,4,5,6,6,7

我的输出数组应该是

1,3,4,5,6,7,0,0,0

我对这个问题的第一次尝试是创建一个交换器，以便在删除元素后将所有 0 推送到列表的末尾，但它似乎不会将零推送到列表的末尾。这是我的代码。

 public void implode(int[] ary) 

    {

        int swapper = -1;

        int[] newARY = new int[ary.length];

        int current = -1;

        for (int i = 0; i < ary.length; i++)

        {

            if (current != ary[i])

            {

            newARY[i] = ary[i];

            current = ary[i];

            }

        }

        for (int i = 0; i < ary.length; i++)

        {

            if (ary[i] == 0)

            {

                if (ary[i + 1] != 0)

                {

                    swapper = ary[i + 1];

                    ary[i] = swapper;

                    ary[i + 1] = 0;

                }

            }

        }

        ary = newARY;

        for (int i = 0; i < newARY.length; i++)

        {

            System.out.print(newARY[i] + " ");

        }

    }

我用它测试的数组是，

 int[] aryIn2 = {1, 1, 2, 3, 4, 4, 5, 6};

但是，当输出内爆阵列时，我收到了这个。

1 0 2 3 4 0 5 6

有什么我想念的吗？

提前致谢。

Answer 1

我观察到的代码有两个问题。1) 您的交换器逻辑在与您之前进行修改的阵列不同的阵列上执行交换2）您需要以冒泡排序的方式拥有此逻辑，即在循环内循环以下是您的方法的有效修改示例代码。我只修改了第二个 for 循环逻辑public void implode(int[] ary) {&nbsp; &nbsp; int swapper = -1;&nbsp; &nbsp; int[] newARY = new int[ary.length];&nbsp; &nbsp; int current = -1;&nbsp; &nbsp; for (int i = 0; i < ary.length; i++) {&nbsp; &nbsp; &nbsp; &nbsp; if (current != ary[i]) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; newARY[i] = ary[i];&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; current = ary[i];&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }&nbsp; &nbsp; for (int i = 0; i < newARY.length - 1; i++) {&nbsp; &nbsp; &nbsp; &nbsp; if (newARY[i] == 0 && newARY[i + 1] != 0) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; for (int j = i; (j + 1) < newARY.length; j++) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; swapper = newARY[j + 1];&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; newARY[j] = swapper;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; newARY[j + 1] = 0;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }&nbsp; &nbsp; for (int i = 0; i < newARY.length; i++) {&nbsp; &nbsp; &nbsp; &nbsp; System.out.print(newARY[i] + " ");&nbsp; &nbsp; }}

Answer 2

不是您问题的答案，但使用（如果可能）Java 流可以缩短您的方式：int[] arr = {1,3,3,4,4,5,6,6,7};// distinctList<Integer> list = Arrays.stream(arr).distinct().boxed().collect(Collectors.toList());// pad with zero'swhile(list.size() < arr.length) {&nbsp; &nbsp; list.add(0);}// displaySystem.out.println(list.stream().map(String::valueOf).collect(Collectors.joining(",")));会输出1,3,4,5,6,7,0,0,0

Answer 3

在第一个循环中：for (int i = 0; i < ary.length; i++) {&nbsp; &nbsp; if (current != ary[i]) {&nbsp; &nbsp; &nbsp; &nbsp; newARY[i] = ary[i];&nbsp; &nbsp; &nbsp; &nbsp; current = ary[i];&nbsp; &nbsp; }}你用重复的值填充newARY元素ary变成0：newARY: 1 0 2 3 4 0 5 6但是，在第二个循环中：&nbsp;for (int i = 0; i < ary.length; i++)&nbsp; &nbsp; {&nbsp; &nbsp; &nbsp; &nbsp; if (ary[i] == 0)&nbsp; &nbsp; &nbsp; &nbsp; {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if (ary[i + 1] != 0)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; swapper = ary[i + 1];&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; ary[i] = swapper;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; ary[i + 1] = 0;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }您正在修改原始ary数组。所以 newARY 没有更新。但是，如果连续有两个以上的 0，您将 0 推到数组末尾的尝试也会失败。而且它也容易受到 ArrayOutOfBoundIndexException 的影响，因为您尝试ary[i+1]不受限制地阅读i将 0 推到数组末尾的一种简单而直接的方法是创建具有非 0 元素的新数组，然后再填充 0：int[] result = new int[ary.lenght];int resultIndex = 0;for (int i = 0; i < newARY.length; i++) {&nbsp; &nbsp; if (newARY[i] != 0) {&nbsp; &nbsp; &nbsp; &nbsp; result[resultIndex++] = newAry[i];&nbsp; &nbsp; }}for (int i = resultIndex; i < newARY.length; i++) {&nbsp; &nbsp; result[i] = 0;}// Print result array提示：使用上述策略，您可以简化您的代码。无需创建直接数组 newARY。只需循环遍历原始数组，将唯一元素推送到结果数组，然后用 0 填充剩余的任何插槽。