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

只是想添加,您也可以使用System.arraycopy:

import static java.lang.System.out;
import static java.lang.System.arraycopy;
import java.lang.reflect.Array;
class Playground {
    @SuppressWarnings("unchecked")
    public static <T>T[] combineArrays(T[] a1, T[] a2) {
        T[] result = (T[]) Array.newInstance(a1.getClass().getComponentType(), a1.length+a2.length);
        arraycopy(a1,0,result,0,a1.length);
        arraycopy(a2,0,result,a1.length,a2.length);
        return result;
    }
    public static void main(String[ ] args) {
        String monthsString = "JANFEBMARAPRMAYJUNJULAUGSEPOCTNOVDEC";
        String[] months = monthsString.split("(?<=\\G.{3})");
        String daysString = "SUNMONTUEWEDTHUFRISAT";
        String[] days = daysString.split("(?<=\\G.{3})");
        for (String m : months) {
            out.println(m);
        }
        out.println("===");
         for (String d : days) {
            out.println(d);
        }
        out.println("===");
        String[] results = combineArrays(months, days);
        for (String r : results) {
            out.println(r);
        }
        out.println("===");
    }
}

其他回答

Object[] obj = {"hi","there"};
Object[] obj2 ={"im","fine","what abt u"};
Object[] obj3 = new Object[obj.length+obj2.length];

for(int i =0;i<obj3.length;i++)
    obj3[i] = (i<obj.length)?obj[i]:obj2[i-obj.length];

使用Java API:

String[] f(String[] first, String[] second) {
    List<String> both = new ArrayList<String>(first.length + second.length);
    Collections.addAll(both, first);
    Collections.addAll(both, second);
    return both.toArray(new String[both.size()]);
}

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

我能找到的最简单的方法如下:


List allFiltersList = Arrays.asList(regularFilters);
allFiltersList.addAll(Arrays.asList(preFiltersArray));
Filter[] mergedFilterArray = (Filter[]) allFiltersList.toArray();

下面是一个简单的方法,它将连接两个数组并返回结果:

public <T> T[] concatenate(T[] a, T[] b) {
    int aLen = a.length;
    int bLen = b.length;

    @SuppressWarnings("unchecked")
    T[] c = (T[]) Array.newInstance(a.getClass().getComponentType(), aLen + bLen);
    System.arraycopy(a, 0, c, 0, aLen);
    System.arraycopy(b, 0, c, aLen, bLen);

    return c;
}

请注意,它不适用于基本数据类型,仅适用于对象类型。

以下稍微复杂一点的版本同时适用于对象数组和基元数组。它通过使用T而不是T[]作为参数类型来实现这一点。

它还可以通过选择最通用的类型作为结果的组件类型来连接两种不同类型的数组。

public static <T> T concatenate(T a, T b) {
    if (!a.getClass().isArray() || !b.getClass().isArray()) {
        throw new IllegalArgumentException();
    }

    Class<?> resCompType;
    Class<?> aCompType = a.getClass().getComponentType();
    Class<?> bCompType = b.getClass().getComponentType();

    if (aCompType.isAssignableFrom(bCompType)) {
        resCompType = aCompType;
    } else if (bCompType.isAssignableFrom(aCompType)) {
        resCompType = bCompType;
    } else {
        throw new IllegalArgumentException();
    }

    int aLen = Array.getLength(a);
    int bLen = Array.getLength(b);

    @SuppressWarnings("unchecked")
    T result = (T) Array.newInstance(resCompType, aLen + bLen);
    System.arraycopy(a, 0, result, 0, aLen);
    System.arraycopy(b, 0, result, aLen, bLen);        

    return result;
}

下面是一个示例:

Assert.assertArrayEquals(new int[] { 1, 2, 3 }, concatenate(new int[] { 1, 2 }, new int[] { 3 }));
Assert.assertArrayEquals(new Number[] { 1, 2, 3f }, concatenate(new Integer[] { 1, 2 }, new Number[] { 3f }));