我需要在Java中连接两个字符串数组。

void f(String[] first, String[] second) {
    String[] both = ???
}

哪种方法最简单?


当前回答

这是我对约阿希姆·绍尔(Joachim Sauer)的concatAll略有改进的版本。它可以在Java5或6上运行,如果运行时可用,可以使用Java6的System.arraycopy。这个方法(IMHO)非常适合Android,因为它在Android<9(没有System.arraycopy)上运行,但如果可能的话,它将使用更快的方法。

  public static <T> T[] concatAll(T[] first, T[]... rest) {
    int totalLength = first.length;
    for (T[] array : rest) {
      totalLength += array.length;
    }
    T[] result;
    try {
      Method arraysCopyOf = Arrays.class.getMethod("copyOf", Object[].class, int.class);
      result = (T[]) arraysCopyOf.invoke(null, first, totalLength);
    } catch (Exception e){
      //Java 6 / Android >= 9 way didn't work, so use the "traditional" approach
      result = (T[]) java.lang.reflect.Array.newInstance(first.getClass().getComponentType(), totalLength);
      System.arraycopy(first, 0, result, 0, first.length);
    }
    int offset = first.length;
    for (T[] array : rest) {
      System.arraycopy(array, 0, result, offset, array.length);
      offset += array.length;
    }
    return result;
  }

其他回答

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];
}

可以编写一个完全通用的版本,甚至可以扩展到连接任意数量的数组。这些版本需要Java 6,因为它们使用Array.copyOf()

这两个版本都避免创建任何中间List对象,并使用System.arraycopy()确保复制大型数组的速度尽可能快。

对于两个阵列,其外观如下:

public static <T> T[] concat(T[] first, T[] second) {
  T[] result = Arrays.copyOf(first, first.length + second.length);
  System.arraycopy(second, 0, result, first.length, second.length);
  return result;
}

对于任意数量的数组(>=1),如下所示:

public static <T> T[] concatAll(T[] first, T[]... rest) {
  int totalLength = first.length;
  for (T[] array : rest) {
    totalLength += array.length;
  }
  T[] result = Arrays.copyOf(first, totalLength);
  int offset = first.length;
  for (T[] array : rest) {
    System.arraycopy(array, 0, result, offset, array.length);
    offset += array.length;
  }
  return result;
}

怎么样:

public String[] combineArray (String[] ... strings) {
    List<String> tmpList = new ArrayList<String>();
    for (int i = 0; i < strings.length; i++)
        tmpList.addAll(Arrays.asList(strings[i]));
    return tmpList.toArray(new String[tmpList.size()]);
}

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]

另一个基于SilverTab的建议,但它支持x个参数,不需要Java6。它也不是通用的,但我确信它可以是通用的。

private byte[] concat(byte[]... args)
{
    int fulllength = 0;
    for (byte[] arrItem : args)
    {
        fulllength += arrItem.length;
    }
    byte[] retArray = new byte[fulllength];
    int start = 0;
    for (byte[] arrItem : args)
    {
        System.arraycopy(arrItem, 0, retArray, start, arrItem.length);
        start += arrItem.length;
    }
    return retArray;
}