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。
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。
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;
}
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;
您可以在两行代码中追加这两个数组。
String[] both = Arrays.copyOf(first, first.length + second.length);
System.arraycopy(second, 0, both, first.length, second.length);
这是一个快速有效的解决方案,适用于原始类型以及所涉及的两个方法都是重载的。
您应该避免使用涉及ArrayList、流等的解决方案,因为这些解决方案需要分配临时内存,没有任何用处。
对于大型数组,应避免for循环,因为这些循环效率不高。内置方法使用速度极快的块复制函数。
或者与心爱的瓜娃:
String[] both = ObjectArrays.concat(first, second, String.class);
此外,基元数组也有一些版本:
布尔型.凹形(第一个,第二个)字节.concat(第一,第二)字符凹面(第一个,第二个)双凹面(第一,第二)短裤.凹形(第一,第二)Ints.concat(第一,第二)长凹面(第一,第二)浮动凹面(第一,第二)