Java数组子 [英] Java array substring

问题描述

如何创建/实例化一个数组等于另一个数组，其中子的大小是未知的子串：

  INT N; //一定数量的衍生别的地方的String [] = GRP元素[I]，以元素[I + N]。
 

解决方案

使用<一个href=\"http://download.oracle.com/javase/6/docs/api/java/util/Arrays.html#copyOfRange%28T%5B%5D,%20int,%20int%29\"><$c$c>Arrays.copyOfRange:

 公共静态＆LT; T＆GT; T [] copyOfRange（T []原创，
                                  从int，
                                  诠释到）
 

  

复制指定的范围内
  指定数组转换成一个新的数组。该
  该范围的初始索引（从）必须
  0之间撒谎， original.length ，
  包括的。在值原[从]
  被放置到的初始元件
  副本（除非从==
  original.length 或从==来）。值
  从在随后的元件
  原始数组被放入
  在副本的后续元素。该
  该范围的最后索引（到），这
  必须大于或等于从，
  可能比 original.length 更大，
  在这种情况下，空被放置在所有的
  复制索引为元素
  大于或等于
   original.length - 从。的长度
  返回的数组将是来 - 从

  
  

所得数组是完全
  同一类原始数组。

在您的情况：

 的String [] = GRP Arrays.copyOfRange（元素，I，I + N）;
 

How can I create/instantiate an array to be equal to the substring of another array, where the size of the substring is unknown:

int n; //some number derived somewhere else

String[] grp = elements[i] to elements[i+n];

解决方案

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 and original.length, inclusive. The value at original[from] is placed into the initial element of the copy (unless from == original.length or from == 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 than original.length, in which case null is placed in all elements of the copy whose index is greater than or equal to original.length - from. The length of the returned array will be to - from.

The resulting array is of exactly the same class as the original array.

In your case:

String[] grp = Arrays.copyOfRange(elements, i, i + n);

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