算法爱好者的另一个答案是：

public static String[] mergeArrays(String[] array1, String[] array2) {
    int totalSize = array1.length + array2.length; // Get total size
    String[] merged = new String[totalSize]; // Create new array
    // Loop over the total size
    for (int i = 0; i < totalSize; i++) {
        if (i < array1.length) // If the current position is less than the length of the first array, take value from first array
            merged[i] = array1[i]; // Position in first array is the current position

        else // If current position is equal or greater than the first array, take value from second array.
            merged[i] = array2[i - array1.length]; // Position in second array is current position minus length of first array.
    }

    return merged;

用法：

String[] array1str = new String[]{"a", "b", "c", "d"}; 
String[] array2str = new String[]{"e", "f", "g", "h", "i"};
String[] listTotalstr = mergeArrays(array1str, array2str);
System.out.println(Arrays.toString(listTotalstr));

结果：

[a, b, c, d, e, f, g, h, i]

我使用下一个方法使用java8连接任意数量的相同类型的数组：

public static <G> G[] concatenate(IntFunction<G[]> generator, G[] ... arrays) {
    int len = arrays.length;
    if (len == 0) {
        return generator.apply(0);
    } else if (len == 1) {
        return arrays[0];
    }
    int pos = 0;
    Stream<G> result = Stream.concat(Arrays.stream(arrays[pos]), Arrays.stream(arrays[++pos]));
    while (pos < len - 1) {
        result = Stream.concat(result, Arrays.stream(arrays[++pos]));
    }
    return result.toArray(generator);
}

用法：

 concatenate(String[]::new, new String[]{"one"}, new String[]{"two"}, new String[]{"three"}) 

or

 concatenate(Integer[]::new, new Integer[]{1}, new Integer[]{2}, new Integer[]{3})

public int[] mergeArrays(int [] a, int [] b) {
    int [] merged = new int[a.length + b.length];
    int i = 0, k = 0, l = a.length;
    int j = a.length > b.length ? a.length : b.length;
    while(i < j) {
        if(k < a.length) {
            merged[k] = a[k];
            k++;
        }
        if((l - a.length) < b.length) {
            merged[l] = b[l - a.length];
            l++;
        }
        i++;
    }
    return merged;
}

这应该是一个衬垫。

public String [] concatenate (final String array1[], final String array2[])
{
    return Stream.concat(Stream.of(array1), Stream.of(array2)).toArray(String[]::new);
}

这一个只适用于int，但想法是通用的

public static int[] junta(int[] v, int[] w) {

int[] junta = new int[v.length + w.length];

for (int i = 0; i < v.length; i++) {            
    junta[i] = v[i];
}

for (int j = v.length; j < junta.length; j++) {
    junta[j] = w[j - v.length];
}

我刚刚发现了这个问题，很抱歉，很晚了，我看到了很多太遥远的答案，使用某些库，使用将数据从数组转换为流并返回到数组等功能。但是，我们只需要使用一个简单的循环，问题就解决了

public String[] concat(String[] firstArr,String[] secondArr){
        //if both is empty just return
        if(firstArr.length==0 && secondArr.length==0)return new String[0];

        String[] res = new String[firstArr.length+secondArr.length];
        int idxFromFirst=0;

        //loop over firstArr, idxFromFirst will be used as starting offset for secondArr
        for(int i=0;i<firstArr.length;i++){
            res[i] = firstArr[i];
            idxFromFirst++;
        }

        //loop over secondArr, with starting offset idxFromFirst (the offset track from first array)
        for(int i=0;i<secondArr.length;i++){
            res[idxFromFirst+i]=secondArr[i];
        }

        return res;
    }

就这样，对吧？他没有说他关心订单或任何事情。这应该是最简单的方法。