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
一个与类型无关的变体(已更新-感谢Volley实例化T):
@SuppressWarnings("unchecked")
public static <T> T[] join(T[]...arrays) {
final List<T> output = new ArrayList<T>();
for(T[] array : arrays) {
output.addAll(Arrays.asList(array));
}
return output.toArray((T[])Array.newInstance(
arrays[0].getClass().getComponentType(), output.size()));
}
其他回答
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]
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!
您可以尝试将其转换为ArrayList,然后使用addAll方法将其转换回数组。
List list = new ArrayList(Arrays.asList(first));
list.addAll(Arrays.asList(second));
String[] both = list.toArray();
使用Java集合
好吧,Java没有提供连接数组的助手方法。然而,自Java5以来,Collections实用程序类引入了addAll(Collection<?super T>c,T…elements)方法。
我们可以创建一个List对象,然后调用该方法两次,将这两个数组添加到列表中。最后,我们将生成的List转换回数组:
static <T> T[] concatWithCollection(T[] array1, T[] array2) {
List<T> resultList = new ArrayList<>(array1.length + array2.length);
Collections.addAll(resultList, array1);
Collections.addAll(resultList, array2);
@SuppressWarnings("unchecked")
//the type cast is safe as the array1 has the type T[]
T[] resultArray = (T[]) Array.newInstance(array1.getClass().getComponentType(), 0);
return resultList.toArray(resultArray);
}
Test
@Test
public void givenTwoStringArrays_whenConcatWithList_thenGetExpectedResult() {
String[] result = ArrayConcatUtil.concatWithCollection(strArray1, strArray2);
assertThat(result).isEqualTo(expectedStringArray);
}