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;
}

其他回答

在Java 8中使用流:

String[] both = Stream.concat(Arrays.stream(a), Arrays.stream(b))
                      .toArray(String[]::new);

或者像这样,使用flatMap:

String[] both = Stream.of(a, b).flatMap(Stream::of)
                      .toArray(String[]::new);

要对泛型类型执行此操作,必须使用反射:

@SuppressWarnings("unchecked")
T[] both = Stream.concat(Arrays.stream(a), Arrays.stream(b)).toArray(
    size -> (T[]) Array.newInstance(a.getClass().getComponentType(), size));

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

用lambda连接一系列紧凑、快速且类型安全的数组

@SafeVarargs
public static <T> T[] concat( T[]... arrays ) {
  return( Stream.of( arrays ).reduce( ( arr1, arr2 ) -> {
      T[] rslt = Arrays.copyOf( arr1, arr1.length + arr2.length );
      System.arraycopy( arr2, 0, rslt, arr1.length, arr2.length );
      return( rslt );
    } ).orElse( null ) );
};

在没有参数的情况下调用时返回null

例如,具有3个阵列:

String[] a = new String[] { "a", "b", "c", "d" };
String[] b = new String[] { "e", "f", "g", "h" };
String[] c = new String[] { "i", "j", "k", "l" };

concat( a, b, c );  // [a, b, c, d, e, f, g, h, i, j, k, l]

“……可能是唯一通用和类型安全的方法”–适用于:

Number[] array1 = { 1, 2, 3 };
Number[] array2 = { 4.0, 5.0, 6.0 };
Number[] array = concat( array1, array2 );  // [1, 2, 3, 4.0, 5.0, 6.0]

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]