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
非Java 8解决方案:
public static int[] combineArrays(int[] a, int[] b) {
int[] c = new int[a.length + b.length];
for (int i = 0; i < a.length; i++) {
c[i] = a[i];
}
for (int j = 0, k = a.length; j < b.length; j++, k++) {
c[k] = b[j];
}
return c;
}
其他回答
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;
}
怎么样:
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()]);
}
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});
}
这里是silvertab解决方案的一个修改,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;
}
注意:请参阅Joachim的Java 6解决方案答案。它不仅消除了警告;它也更短,更高效,更容易阅读!
算法爱好者的另一个答案是:
public static String[] mergeArrays(String[] array1, String[] array2) {
int totalSize = array1.length + array2.length; // Get total size
String[] merged = new String[totalSize]; // Create new array
// Loop over the total size
for (int i = 0; i < totalSize; i++) {
if (i < array1.length) // If the current position is less than the length of the first array, take value from first array
merged[i] = array1[i]; // Position in first array is the current position
else // If current position is equal or greater than the first array, take value from second array.
merged[i] = array2[i - array1.length]; // Position in second array is current position minus length of first array.
}
return merged;
用法:
String[] array1str = new String[]{"a", "b", "c", "d"};
String[] array2str = new String[]{"e", "f", "g", "h", "i"};
String[] listTotalstr = mergeArrays(array1str, array2str);
System.out.println(Arrays.toString(listTotalstr));
结果:
[a, b, c, d, e, f, g, h, i]