Object[] obj = {"hi","there"};
Object[] obj2 ={"im","fine","what abt u"};
Object[] obj3 = new Object[obj.length+obj2.length];

for(int i =0;i<obj3.length;i++)
    obj3[i] = (i<obj.length)?obj[i]:obj2[i-obj.length];

您可以尝试将其转换为ArrayList，然后使用addAll方法将其转换回数组。

List list = new ArrayList(Arrays.asList(first));
  list.addAll(Arrays.asList(second));
  String[] both = list.toArray();

public String[] concat(String[]... arrays)
{
    int length = 0;
    for (String[] array : arrays) {
        length += array.length;
    }
    String[] result = new String[length];
    int destPos = 0;
    for (String[] array : arrays) {
        System.arraycopy(array, 0, result, destPos, array.length);
        destPos += array.length;
    }
    return result;
}

String [] both = new ArrayList<String>(){{addAll(Arrays.asList(first)); addAll(Arrays.asList(second));}}.toArray(new String[0]);

我能找到的最简单的方法如下：

List allFiltersList = Arrays.asList(regularFilters);
allFiltersList.addAll(Arrays.asList(preFiltersArray));
Filter[] mergedFilterArray = (Filter[]) allFiltersList.toArray();

看看这个优雅的解决方案（如果您需要除char以外的其他类型，请更改它）：

private static void concatArrays(char[] destination, char[]... sources) {
    int currPos = 0;
    for (char[] source : sources) {
        int length = source.length;
        System.arraycopy(source, 0, destination, currPos, length);
        currPos += length;
    }
}

您可以连接每个数组计数。