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
/**
* With Java Streams
* @param first First Array
* @param second Second Array
* @return Merged Array
*/
String[] mergeArrayOfStrings(String[] first, String[] second) {
return Stream.concat(Arrays.stream(first), Arrays.stream(second)).toArray(String[]::new);
}
其他回答
这是可行的,但您需要插入自己的错误检查。
public class StringConcatenate {
public static void main(String[] args){
// Create two arrays to concatenate and one array to hold both
String[] arr1 = new String[]{"s","t","r","i","n","g"};
String[] arr2 = new String[]{"s","t","r","i","n","g"};
String[] arrBoth = new String[arr1.length+arr2.length];
// Copy elements from first array into first part of new array
for(int i = 0; i < arr1.length; i++){
arrBoth[i] = arr1[i];
}
// Copy elements from second array into last part of new array
for(int j = arr1.length;j < arrBoth.length;j++){
arrBoth[j] = arr2[j-arr1.length];
}
// Print result
for(int k = 0; k < arrBoth.length; k++){
System.out.print(arrBoth[k]);
}
// Additional line to make your terminal look better at completion!
System.out.println();
}
}
它可能不是最有效的,但除了Java自己的API之外,它不依赖其他任何东西。
这一个只适用于int,但想法是通用的
public static int[] junta(int[] v, int[] w) {
int[] junta = new int[v.length + w.length];
for (int i = 0; i < v.length; i++) {
junta[i] = v[i];
}
for (int j = v.length; j < junta.length; j++) {
junta[j] = w[j - v.length];
}
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]
一种简单但低效的方法(不包括泛型):
ArrayList baseArray = new ArrayList(Arrays.asList(array1));
baseArray.addAll(Arrays.asList(array2));
String concatenated[] = (String []) baseArray.toArray(new String[baseArray.size()]);
这里是silvertab编写的伪代码解决方案的工作代码中的一个可能实现。
谢谢silvertab!
public class Array {
public static <T> T[] concat(T[] a, T[] b, ArrayBuilderI<T> builder) {
T[] c = builder.build(a.length + b.length);
System.arraycopy(a, 0, c, 0, a.length);
System.arraycopy(b, 0, c, a.length, b.length);
return c;
}
}
接下来是构建器界面。
注意:构建器是必要的,因为在java中不可能这样做
新T[尺寸]
由于通用类型擦除:
public interface ArrayBuilderI<T> {
public T[] build(int size);
}
这里是一个实现接口的具体构建器,构建一个整数数组:
public class IntegerArrayBuilder implements ArrayBuilderI<Integer> {
@Override
public Integer[] build(int size) {
return new Integer[size];
}
}
最后是应用程序/测试:
@Test
public class ArrayTest {
public void array_concatenation() {
Integer a[] = new Integer[]{0,1};
Integer b[] = new Integer[]{2,3};
Integer c[] = Array.concat(a, b, new IntegerArrayBuilder());
assertEquals(4, c.length);
assertEquals(0, (int)c[0]);
assertEquals(1, (int)c[1]);
assertEquals(2, (int)c[2]);
assertEquals(3, (int)c[3]);
}
}