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

这是算盘常用的密码。

String[] a = {"a", "b", "c"};
String[] b = {"1", "2", "3"};
String[] c = N.concat(a, b); // c = ["a", "b", "c", "1", "2", "3"]

// N.concat(...) is null-safety.
a = null;
c = N.concat(a, b); // c = ["1", "2", "3"]

其他回答

ArrayList<String> both = new ArrayList(Arrays.asList(first));
both.addAll(Arrays.asList(second));

both.toArray(new String[0]);

我测试了以下代码,工作正常

我还使用了library:org.apache.mons.lang.ArrayUtils

public void testConcatArrayString(){
    String[] a = null;
    String[] b = null;
    String[] c = null;
    a = new String[] {"1","2","3","4","5"};
    b = new String[] {"A","B","C","D","E"};

    c = (String[]) ArrayUtils.addAll(a, b);
    if(c!=null){
        for(int i=0; i<c.length; i++){
            System.out.println("c[" + (i+1) + "] = " + c[i]);
        }
    }
}

当做

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;
}

允许连接多个阵列的简单变体:

public static String[] join(String[]...arrays) {

    final List<String> output = new ArrayList<String>();

    for(String[] array : arrays) {
        output.addAll(Arrays.asList(array));
    }

    return output.toArray(new String[output.size()]);
}

您可以尝试连接多个数组的方法:

public static <T> T[] concatMultipleArrays(T[]... arrays)
{
   int length = 0;
   for (T[] array : arrays)
   {
      length += array.length;
   }
   T[] result = (T[]) Array.newInstance(arrays.getClass().getComponentType(), length) ;

   length = 0;
   for (int i = 0; i < arrays.length; i++)
   {
      System.arraycopy(arrays[i], 0, result, length, arrays[i].length);
      length += arrays[i].length;
   }

   return result;
}