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?
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);
}
其他回答
你可以试试这个
public static Object[] addTwoArray(Object[] objArr1, Object[] objArr2){
int arr1Length = objArr1!=null && objArr1.length>0?objArr1.length:0;
int arr2Length = objArr2!=null && objArr2.length>0?objArr2.length:0;
Object[] resutlentArray = new Object[arr1Length+arr2Length];
for(int i=0,j=0;i<resutlentArray.length;i++){
if(i+1<=arr1Length){
resutlentArray[i]=objArr1[i];
}else{
resutlentArray[i]=objArr2[j];
j++;
}
}
return resutlentArray;
}
你可以键入你的数组!!!
使用高性能System.arraycopy而不需要@SuppressWarnings注释的通用静态版本:
public static <T> T[] arrayConcat(T[] a, T[] b) {
T[] both = Arrays.copyOf(a, a.length + b.length);
System.arraycopy(b, 0, both, a.length, b.length);
return both;
}
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!
Object[] mixArray(String[] a, String[] b)
String[] s1 = a;
String[] s2 = b;
Object[] result;
List<String> input = new ArrayList<String>();
for (int i = 0; i < s1.length; i++)
{
input.add(s1[i]);
}
for (int i = 0; i < s2.length; i++)
{
input.add(s2[i]);
}
result = input.toArray();
return result;