你看起来似乎把事情弄得太复杂了，因为一个简单的嵌套for循环可能就是你所需要的。外层循环遍历数组，直到但不包括最后一个数字，内层循环在数组中的一个项到下一个项之间循环，从arr[i] +到arr[i + 1]。注意，如果两个数组项是连续的，内层循环就不会“循环”。并且因此不需要IF块：

public class MissingNumber {
    public static void main(String[] args) {
        int[] arr = {12, 13, 17, 18, 20, 21, 24, 25, 26};

        System.out.println("missing numbers");
        for (int i = 0; i < arr.length - 1; i++) {
            for (int j = arr[i] + 1; j < arr[i + 1]; j++) {
                System.out.println("" + j);
            }
        }
    }
}

输出量：

missing numbers
14
15
16
19
22
23

就是这样

您可以使用List<Integer>来避免重复相同的计算。

static List<Integer> missingNumbers(int[] a) {
    List<Integer> missingNumbers = new ArrayList<>();
    for (int i = 1, length = a.length; i < length; ++i)
        if (a[i - 1] + 1 < a[i])
            for (int j = a[i - 1] + 1; j < a[i]; ++j)
                missingNumbers.add(j);
    return missingNumbers;
}

public static void main(String[] args) {
    int[] a = {12, 13, 17, 18, 20, 21, 24, 25, 26};
    List<Integer> missingNumbers = missingNumbers(a);
    if (missingNumbers.size() == 1)
        System.out.println("There is only one missing number:");
    else
        System.out.println("There are " + missingNumbers.size() + " missing numbers:");
    for (int n : missingNumbers)
        System.out.println(n);
}

输出：

There are 6 missing numbers:
14
15
16
19
22
23

int[] arr = { 12, 13, 17, 18, 20, 21, 24, 25, 26 };
System.out.println("Array Source:");
System.out.println(Arrays.toString(arr) + "\n");
System.out.println("Missing number(s):");
int nextNumber = arr[0] + 1;
for (int i = 1; i < arr.length; i++) {
    while (arr[i] > nextNumber) {
        System.out.println(nextNumber);
        nextNumber++;
    }
    nextNumber++;
}

要用从min到max的所有数字填充数组，您可以写入

int[] shouldBe = IntStream.range(min, max + 1).toArray();

在这种特殊情况下

int[] shouldBe = IntStream.range(arr[0], arr[arr.length - 1] + 1).toArray();