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
这是字符串数组的转换函数:
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;
}
其他回答
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]
void f(String[] first, String[] second) {
String[] both = new String[first.length+second.length];
for(int i=0;i<first.length;i++)
both[i] = first[i];
for(int i=0;i<second.length;i++)
both[first.length + i] = second[i];
}
这一个在不了解任何其他类/库等的情况下工作。它适用于任何数据类型。只需将String替换为int、double或char等任意值。它的工作效率很高。
我能找到的最简单的方法如下:
List allFiltersList = Arrays.asList(regularFilters);
allFiltersList.addAll(Arrays.asList(preFiltersArray));
Filter[] mergedFilterArray = (Filter[]) allFiltersList.toArray();
一个100%旧的java和没有System.arraycopy的解决方案(例如GWT客户端中不可用):
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;
}
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});
}