Java数组子串
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Java array substring
提问by Will
How can I create/instantiate an array to be equal to the substring of another array, where the size of the substring is unknown:
如何创建/实例化一个数组以等于另一个数组的子字符串，其中子字符串的大小未知：
int n; //some number derived somewhere else
String[] grp = elements[i] to elements[i+n];
采纳答案by NPE
Use Arrays.copyOfRange:
public static <T> T[] copyOfRange(T[] original,
int from,
int to)
Copies the specified range of the specified array into a new array. The initial index of the range (
from) must lie between zero andoriginal.length, inclusive. The value atoriginal[from]is placed into the initial element of the copy (unlessfrom == original.lengthorfrom == to). Values from subsequent elements in the original array are placed into subsequent elements in the copy. The final index of the range (to), which must be greater than or equal to from, may be greater thanoriginal.length, in which case null is placed in all elements of the copy whose index is greater than or equal tooriginal.length - from. The length of the returned array will beto - from.The resulting array is of exactly the same class as the original array.
将指定数组的指定范围复制到新数组中。范围 (
from)的初始索引必须介于零和original.length,之间。值 atoriginal[from]放置在副本的初始元素中（除非from == original.length或from == to）。来自原始数组中后续元素的值被放置到副本中的后续元素中。范围 (to)的最终索引（必须大于或等于 from）可能大于original.length，在这种情况下，将 null 放置在索引大于或等于 的副本的所有元素中original.length - from。返回数组的长度将为to - from.生成的数组与原始数组的类完全相同。
In your case:
在你的情况下：
String[] grp = Arrays.copyOfRange(elements, i, i + n);
回答by jjnguy
回答by Joachim Sauer
Use Arrays.copyOfRange():
String[] grp = Arrays.copyOfRange(grp, i, i+n);
As the name implies grpwill be a copyof the original array and not a view into it. You can't have views into the array, for that you'd need to use a collection. Generally speaking collections are the more powerful and flexible, high-level alternative to arrays.
顾名思义，grp将是原始数组的副本，而不是其中的视图。您不能在数组中查看视图，为此您需要使用集合。一般来说，集合是数组的更强大、更灵活、更高级的替代品。
回答by JVM
To get the size of the array, you would do
要获得数组的大小，你会做
String [] grp = new String[n + 1];//inclusive
Then, all you have to do is copy the elements over:
然后，您所要做的就是复制元素：
for(int x = 0;x < n + 1;x++)
{
grp[x] = elements[i + x];//I'm assuming you have "i" defined somewhere
}
