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
这是算盘常用的密码。
String[] a = {"a", "b", "c"};
String[] b = {"1", "2", "3"};
String[] c = N.concat(a, b); // c = ["a", "b", "c", "1", "2", "3"]
// N.concat(...) is null-safety.
a = null;
c = N.concat(a, b); // c = ["1", "2", "3"]
其他回答
ArrayList<String> both = new ArrayList(Arrays.asList(first));
both.addAll(Arrays.asList(second));
both.toArray(new String[0]);
我测试了以下代码,工作正常
我还使用了library:org.apache.mons.lang.ArrayUtils
public void testConcatArrayString(){
String[] a = null;
String[] b = null;
String[] c = null;
a = new String[] {"1","2","3","4","5"};
b = new String[] {"A","B","C","D","E"};
c = (String[]) ArrayUtils.addAll(a, b);
if(c!=null){
for(int i=0; i<c.length; i++){
System.out.println("c[" + (i+1) + "] = " + c[i]);
}
}
}
当做
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;
}
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()]);
}
您可以尝试连接多个数组的方法:
public static <T> T[] concatMultipleArrays(T[]... arrays)
{
int length = 0;
for (T[] array : arrays)
{
length += array.length;
}
T[] result = (T[]) Array.newInstance(arrays.getClass().getComponentType(), length) ;
length = 0;
for (int i = 0; i < arrays.length; i++)
{
System.arraycopy(arrays[i], 0, result, length, arrays[i].length);
length += arrays[i].length;
}
return result;
}