如何在Java中连接两个数组?
我需要在Java中连接两个如何在Java中连接两个数组?,java,arrays,concatenation,add,Java,Arrays,Concatenation,Add,我需要在Java中连接两个String数组 void f(String[] first, String[] second) { String[] both = ??? } 最简单的方法是什么?这里有一个简单的方法,可以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("
String
数组
void f(String[] first, String[] second) {
String[] both = ???
}
最简单的方法是什么?这里有一个简单的方法,可以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;
}
以下是一个例子:
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 }));
我从旧的Apache Commons Lang库中找到了一个单行解决方案。
代码: 该库有一个数组包装器类,它为数组配备了方便的方法,如连接
import static fj.data.Array.array;
…然后
Array<String> both = array(first).append(array(second));
仅使用Java自己的API:
String[] join(String[]... arrays) {
// calculate size of target array
int size = 0;
for (String[] array : arrays) {
size += array.length;
}
// create list of appropriate size
java.util.List list = new java.util.ArrayList(size);
// add arrays
for (String[] array : arrays) {
list.addAll(java.util.Arrays.asList(array));
}
// create and return final array
return list.toArray(new String[size]);
}
现在,这段代码不是最有效的,但它只依赖于标准java类,并且易于理解。它适用于任意数量的字符串[](甚至是零数组)。这里是silvertab解决方案的一个改编版本,并对泛型进行了改进:
static <T> T[] concat(T[] a, T[] b) {
final int alen = a.length;
final int blen = b.length;
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;
}
static T[]concat(T[]a,T[]b){
最终int alen=a.长度;
最终整数blen=b.长度;
最终的T[]结果=(T[])java.lang.reflect.Array。
newInstance(a.getClass().getComponentType(),alen+blen);
数组复制(a,0,result,0,alen);
数组副本(b,0,result,alen,blen);
返回结果;
}
注意:有关Java 6解决方案,请参阅。它不仅消除了警告;它也更短,更高效,更容易阅读 我最近一直在与过度的内存旋转问题作斗争。如果已知a和/或b通常为空,下面是silvertab代码的另一个改编版本(也已通用):
private static T[]concatOrReturnSame(T[]a,T[]b){
最终int alen=a.长度;
最终整数blen=b.长度;
如果(alen==0){
返回b;
}
如果(blen==0){
返回a;
}
最终的T[]结果=(T[])java.lang.reflect.Array。
newInstance(a.getClass().getComponentType(),alen+blen);
数组复制(a,0,result,0,alen);
数组副本(b,0,result,alen,blen);
返回结果;
}
编辑:这篇文章的前一个版本指出,像这样的阵列重复使用应该被清楚地记录下来。正如Maarten在评论中指出的,一般来说,最好删除if语句,这样就不需要文档。但话说回来,这些if语句首先是这个特定优化的全部要点。我将把这个答案留在这里,但要小心 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()]);
}
String[]f(String[]first,String[]second){
列出两者=新的ArrayList(first.length+second.length);
Collections.addAll(两个,第一个);
集合。addAll(两个,第二个);
返回both.toArray(新字符串[both.size()]);
}
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();
}
}
公共类串接{
公共静态void main(字符串[]args){
//创建两个要连接的数组和一个同时容纳这两个数组的数组
字符串[]arr1=新字符串[]{“s”、“t”、“r”、“i”、“n”、“g”};
字符串[]arr2=新字符串[]{“s”、“t”、“r”、“i”、“n”、“g”};
String[]arrleath=新字符串[arr1.length+arr2.length];
//将元素从第一个数组复制到新数组的第一部分
for(int i=0;i
它可能不是最有效的,但它只依赖于Java自己的API。可以编写一个完全通用的版本,甚至可以扩展到连接任意数量的数组。这些版本需要Java 6,因为它们使用 这两个版本都避免创建任何中间
列表
对象,并使用System.arraycopy()
确保尽可能快地复制大型数组
public static <T> T[] concatAll(T[] first, T[]... rest) {
int totalLength = first.length;
for (T[] array : rest) {
totalLength += array.length;
}
T[] result;
try {
Method arraysCopyOf = Arrays.class.getMethod("copyOf", Object[].class, int.class);
result = (T[]) arraysCopyOf.invoke(null, first, totalLength);
} catch (Exception e){
//Java 6 / Android >= 9 way didn't work, so use the "traditional" approach
result = (T[]) java.lang.reflect.Array.newInstance(first.getClass().getComponentType(), totalLength);
System.arraycopy(first, 0, result, 0, first.length);
}
int offset = first.length;
for (T[] array : rest) {
System.arraycopy(array, 0, result, offset, array.length);
offset += array.length;
}
return result;
}
对于两个阵列,它如下所示:
public static <T> T[] concat(T[] first, T[] second) {
T[] result = Arrays.copyOf(first, first.length + second.length);
System.arraycopy(second, 0, result, first.length, second.length);
return result;
}
public static <T> T[] concatAll(T[] first, T[]... rest) {
int totalLength = first.length;
for (T[] array : rest) {
totalLength += array.length;
}
T[] result = Arrays.copyOf(first, totalLength);
int offset = first.length;
for (T[] array : rest) {
System.arraycopy(array, 0, result, offset, array.length);
offset += array.length;
}
return result;
}
String [] arg1 = new String{"a","b","c"};
String [] arg2 = new String{"x","y","z"};
ArrayList<String> temp = new ArrayList<String>();
temp.addAll(Arrays.asList(arg1));
temp.addAll(Arrays.asList(arg2));
String [] concatedArgs = temp.toArray(new String[arg1.length+arg2.length]);
publicstatict[]concat(T[]first,T[]second){
T[]result=Arrays.copyOf(first,first.length+second.length);
System.arraycopy(秒,0,结果,first.length,second.length);
返回结果;
}
对于任意数量的数组(>=1),它如下所示:
public static <T> T[] concat(T[] first, T[] second) {
T[] result = Arrays.copyOf(first, first.length + second.length);
System.arraycopy(second, 0, result, first.length, second.length);
return result;
}
public static <T> T[] concatAll(T[] first, T[]... rest) {
int totalLength = first.length;
for (T[] array : rest) {
totalLength += array.length;
}
T[] result = Arrays.copyOf(first, totalLength);
int offset = first.length;
for (T[] array : rest) {
System.arraycopy(array, 0, result, offset, array.length);
offset += array.length;
}
return result;
}
String [] arg1 = new String{"a","b","c"};
String [] arg2 = new String{"x","y","z"};
ArrayList<String> temp = new ArrayList<String>();
temp.addAll(Arrays.asList(arg1));
temp.addAll(Arrays.asList(arg2));
String [] concatedArgs = temp.toArray(new String[arg1.length+arg2.length]);
publicstatict[]目录(T[]优先,T[]剩余){
int totalength=第一个长度;
for(T[]数组:rest){
totalLength+=数组长度;
}
T[]结果=数组.copyOf(第一个,总长度);
int offset=first.length;
for(T[]数组:rest){
System.arraycopy(数组,0,结果,偏移量,数组.长度);
偏移量+=数组长度;
}
返回结果;
}
一个简单的变体,允许连接多个阵列:
public static String[] join(String[]...arrays) {
final List<String> output = new ArrayList<String>();
for(String[] array : arrays) {
output.addAll(Arrays.asList(array));
}
return output.toArray(new String[output.size()]);
}
公共静态字符串[]连接(字符串[]…数组){
最终列表输出=新的ArrayList();
for(字符串[]数组:数组){
output.addAll(Arrays.asList(array));
}
返回output.toArray(新字符串[output.size()]);
}
或与心爱的人:
此外,基本体数组还有以下版本:
Booleans.concat(第一,第二)
Bytes.concat(第一,第二)
Chars.concat(第一,第二)
double.concat(第一,第二)
Shorts.concat(第一,第二)
Ints.concat(第一,第二)
Longs.concat(第一,第二)
float.concat(第一,第二)
public String[] mergeArrays(String[] mainArray, String[] addArray) {
String[] finalArray = new String[mainArray.length + addArray.length];
System.arraycopy(mainArray, 0, finalArray, 0, mainArray.length);
System.arraycopy(addArray, 0, finalArray, mainArray.length, addArray.length);
return finalArray;
}
一种解决方案100%旧java和不带
系统。arraycopy
(例如,GWT客户端中不提供):
呵呵
public static <T> T[] concatAll(T[] first, T[]... rest) {
int totalLength = first.length;
for (T[] array : rest) {
totalLength += array.length;
}
T[] result;
try {
Method arraysCopyOf = Arrays.class.getMethod("copyOf", Object[].class, int.class);
result = (T[]) arraysCopyOf.invoke(null, first, totalLength);
} catch (Exception e){
//Java 6 / Android >= 9 way didn't work, so use the "traditional" approach
result = (T[]) java.lang.reflect.Array.newInstance(first.getClass().getComponentType(), totalLength);
System.arraycopy(first, 0, result, 0, first.length);
}
int offset = first.length;
for (T[] array : rest) {
System.arraycopy(array, 0, result, offset, array.length);
offset += array.length;
}
return result;
}
public String[] mergeArrays(String[] mainArray, String[] addArray) {
String[] finalArray = new String[mainArray.length + addArray.length];
System.arraycopy(mainArray, 0, finalArray, 0, mainArray.length);
System.arraycopy(addArray, 0, finalArray, mainArray.length, addArray.length);
return finalArray;
}
static String[] concat(String[]... arrays) {
int length = 0;
for (String[] array : arrays) {
length += array.length;
}
String[] result = new String[length];
int pos = 0;
for (String[] array : arrays) {
for (String element : array) {
result[pos] = element;
pos++;
}
}
return result;
}
public static class Array {
public static <T> T[] concat(T[]... arrays) {
ArrayList<T> al = new ArrayList<T>();
for (T[] one : arrays)
Collections.addAll(al, one);
return (T[]) al.toArray(arrays[0].clone());
}
}
@SuppressWarnings("unchecked")
public static <T> T[] concat(T[]... inputArrays) {
if(inputArrays.length < 2) {
throw new IllegalArgumentException("inputArrays must contain at least 2 arrays");
}
for(int i = 0; i < inputArrays.length; i++) {
if(inputArrays[i] == null) {
throw new IllegalArgumentException("inputArrays[" + i + "] is null");
}
}
int totalLength = 0;
for(T[] array : inputArrays) {
totalLength += array.length;
}
T[] result = (T[]) Array.newInstance(inputArrays[0].getClass().getComponentType(), totalLength);
int offset = 0;
for(T[] array : inputArrays) {
System.arraycopy(array, 0, result, offset, array.length);
offset += array.length;
}
return result;
}
String [] arg1 = new String{"a","b","c"};
String [] arg2 = new String{"x","y","z"};
ArrayList<String> temp = new ArrayList<String>();
temp.addAll(Arrays.asList(arg1));
temp.addAll(Arrays.asList(arg2));
String [] concatedArgs = temp.toArray(new String[arg1.length+arg2.length]);
private static <T> T[] addAll(final T[] f, final T...o){
return new AbstractList<T>(){
@Override
public T get(int i) {
return i>=f.length ? o[i - f.length] : f[i];
}
@Override
public int size() {
return f.length + o.length;
}
}.toArray(f);
}
ArrayList<String> both = new ArrayList(Arrays.asList(first));
both.addAll(Arrays.asList(second));
both.toArray(new String[0]);
String[] both = Stream.concat(Arrays.stream(a), Arrays.stream(b))
.toArray(String[]::new);
String[] both = Stream.of(a, b).flatMap(Stream::of)
.toArray(String[]::new);
@SuppressWarnings("unchecked")
T[] both = Stream.concat(Arrays.stream(a), Arrays.stream(b)).toArray(
size -> (T[]) Array.newInstance(a.getClass().getComponentType(), size));
public String[] combineArray (String[] ... strings) {
List<String> tmpList = new ArrayList<String>();
for (int i = 0; i < strings.length; i++)
tmpList.addAll(Arrays.asList(strings[i]));
return tmpList.toArray(new String[tmpList.size()]);
}
List list = new ArrayList(Arrays.asList(first));
list.addAll(Arrays.asList(second));
String[] both = list.toArray();
public String[] concatString(String[] a, String[] b){
Stream<String> streamA = Arrays.stream(a);
Stream<String> streamB = Arrays.stream(b);
return Stream.concat(streamA, streamB).toArray(String[]::new);
}
public static String[] combineString(String[] first, String[] second){
int length = first.length + second.length;
String[] result = new String[length];
System.arraycopy(first, 0, result, 0, first.length);
System.arraycopy(second, 0, result, first.length, second.length);
return result;
}
public static int[] combineInt(int[] a, int[] b){
int length = a.length + b.length;
int[] result = new int[length];
System.arraycopy(a, 0, result, 0, a.length);
System.arraycopy(b, 0, result, a.length, b.length);
return result;
}
public static void main(String[] args) {
String [] first = {"a", "b", "c"};
String [] second = {"d", "e"};
String [] joined = combineString(first, second);
System.out.println("concatenated String array : " + Arrays.toString(joined));
int[] array1 = {101,102,103,104};
int[] array2 = {105,106,107,108};
int[] concatenateInt = combineInt(array1, array2);
System.out.println("concatenated Int array : " + Arrays.toString(concatenateInt));
}
}
String[] both = Arrays.copyOf(first, first.length + second.length);
System.arraycopy(second, 0, both, first.length, second.length);
private static String[] concatArrays(final String[]... arrays) {
return Arrays.stream(arrays)
.flatMap(Arrays::stream)
.toArray(String[]::new);
}
public String [] concatenate (final String array1[], final String array2[])
{
return Stream.concat(Stream.of(array1), Stream.of(array2)).toArray(String[]::new);
}
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;
}