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
简单一点怎么样
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());
}
}
只需执行Array.concat(arr1,arr2)。只要arr1和arr2是相同类型的,这将为您提供另一个包含这两个数组的相同类型的数组。
其他回答
我看到许多带有公共静态T[]concat(T[]a,T[]b){}等签名的通用答案,但据我所知,这些答案只适用于Object数组,而不适用于基元数组。下面的代码既适用于对象数组,也适用于基元数组,使其更通用。。。
public static <T> T concat(T a, T b) {
//Handles both arrays of Objects and primitives! E.g., int[] out = concat(new int[]{6,7,8}, new int[]{9,10});
//You get a compile error if argument(s) not same type as output. (int[] in example above)
//You get a runtime error if output type is not an array, i.e., when you do something like: int out = concat(6,7);
if (a == null && b == null) return null;
if (a == null) return b;
if (b == null) return a;
final int aLen = Array.getLength(a);
final int bLen = Array.getLength(b);
if (aLen == 0) return b;
if (bLen == 0) return a;
//From here on we really need to concatenate!
Class componentType = a.getClass().getComponentType();
final T result = (T)Array.newInstance(componentType, aLen + bLen);
System.arraycopy(a, 0, result, 0, aLen);
System.arraycopy(b, 0, result, aLen, bLen);
return result;
}
public static void main(String[] args) {
String[] out1 = concat(new String[]{"aap", "monkey"}, new String[]{"rat"});
int[] out2 = concat(new int[]{6,7,8}, new int[]{9,10});
}
public int[] mergeArrays(int [] a, int [] b) {
int [] merged = new int[a.length + b.length];
int i = 0, k = 0, l = a.length;
int j = a.length > b.length ? a.length : b.length;
while(i < j) {
if(k < a.length) {
merged[k] = a[k];
k++;
}
if((l - a.length) < b.length) {
merged[l] = b[l - a.length];
l++;
}
i++;
}
return merged;
}
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;
允许连接多个阵列的简单变体:
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()]);
}
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。