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;
}

这里是silvertab解决方案的一个修改，对泛型进行了改进：

static <T> T[] concat(T[] a, T[] b) {
    final int alen = a.length;
    final int blen = b.length;
    final T[] result = (T[]) java.lang.reflect.Array.
            newInstance(a.getClass().getComponentType(), alen + blen);
    System.arraycopy(a, 0, result, 0, alen);
    System.arraycopy(b, 0, result, alen, blen);
    return result;
}

注意：请参阅Joachim的Java 6解决方案答案。它不仅消除了警告；它也更短，更高效，更容易阅读！

FunctionalJava库有一个数组包装器类，它为数组提供了方便的方法，如串联。

import static fj.data.Array.array;

…然后

Array<String> both = array(first).append(array(second));

要取回展开的数组，请调用

String[] s = both.array();

这是字符串数组的转换函数：

public String[] mergeArrays(String[] mainArray, String[] addArray) {
    String[] finalArray = new String[mainArray.length + addArray.length];
    System.arraycopy(mainArray, 0, finalArray, 0, mainArray.length);
    System.arraycopy(addArray, 0, finalArray, mainArray.length, addArray.length);

    return finalArray;
}

我认为泛型的最佳解决方案是：

/* This for non primitive types */
public static <T> T[] concatenate (T[]... elements) {

    T[] C = null;
    for (T[] element: elements) {
        if (element==null) continue;
        if (C==null) C = (T[]) Array.newInstance(element.getClass().getComponentType(), element.length);
        else C = resizeArray(C, C.length+element.length);

        System.arraycopy(element, 0, C, C.length-element.length, element.length);
    }

    return C;
}

/**
 * as far as i know, primitive types do not accept generics 
 * http://stackoverflow.com/questions/2721546/why-dont-java-generics-support-primitive-types
 * for primitive types we could do something like this:
 * */
public static int[] concatenate (int[]... elements){
    int[] C = null;
    for (int[] element: elements) {
        if (element==null) continue;
        if (C==null) C = new int[element.length];
        else C = resizeArray(C, C.length+element.length);

        System.arraycopy(element, 0, C, C.length-element.length, element.length);
    }
    return C;
}

private static <T> T resizeArray (T array, int newSize) {
    int oldSize =
            java.lang.reflect.Array.getLength(array);
    Class elementType =
            array.getClass().getComponentType();
    Object newArray =
            java.lang.reflect.Array.newInstance(
                    elementType, newSize);
    int preserveLength = Math.min(oldSize, newSize);
    if (preserveLength > 0)
        System.arraycopy(array, 0,
                newArray, 0, preserveLength);
    return (T) newArray;
}

我最近一直在与过度的记忆循环作斗争。如果已知a和/或b通常是空的，这里是silvertab代码的另一种修改（也被通用化）：

private static <T> T[] concatOrReturnSame(T[] a, T[] b) {
    final int alen = a.length;
    final int blen = b.length;
    if (alen == 0) {
        return b;
    }
    if (blen == 0) {
        return a;
    }
    final T[] result = (T[]) java.lang.reflect.Array.
            newInstance(a.getClass().getComponentType(), alen + blen);
    System.arraycopy(a, 0, result, 0, alen);
    System.arraycopy(b, 0, result, alen, blen);
    return result;
}

编辑：这篇文章的前一个版本指出，像这样的数组重用应该清楚地记录下来。正如Maarten在评论中指出的那样，一般来说，最好删除if语句，这样就不需要文档了。但话说回来，那些if语句首先就是这个特定优化的要点。我会在这里留下这个答案，但要小心！