35

是否可以在不迭代整个数组的情况下将字符串添加到字符串数组的开头。

4

10 回答 10

21

做到这一点的唯一方法是维护一个环形缓冲区。即你有一个计数器，它记住开始在哪里，你移动它而不是移动数组中的所有条目。这仅适用于您重新定义“开始”的含义。

请参阅具有三个字段的ArrayDeque的源代码

   86       /**
   87        * The array in which the elements of the deque are stored.
   88        * The capacity of the deque is the length of this array, which is
   89        * always a power of two. The array is never allowed to become
   90        * full, except transiently within an addX method where it is
   91        * resized (see doubleCapacity) immediately upon becoming full,
   92        * thus avoiding head and tail wrapping around to equal each
   93        * other.  We also guarantee that all array cells not holding
   94        * deque elements are always null.
   95        */
   96       private transient E[] elements;
   97   
   98       /**
   99        * The index of the element at the head of the deque (which is the
  100        * element that would be removed by remove() or pop()); or an
  101        * arbitrary number equal to tail if the deque is empty.
  102        */
  103       private transient int head;
  104   
  105       /**
  106        * The index at which the next element would be added to the tail
  107        * of the deque (via addLast(E), add(E), or push(E)).
  108        */
  109       private transient int tail;

所以添加到开始就像这样

  224       public void addFirst(E e) {
  225           if (e == null)
  226               throw new NullPointerException();
  227           elements[head = (head - 1) & (elements.length - 1)] = e;
  228           if (head == tail)
  229               doubleCapacity();
  230       }


  312       /**
  313        * @throws NoSuchElementException {@inheritDoc}
  314        */
  315       public E getFirst() {
  316           E x = elements[head];
  317           if (x == null)
  318               throw new NoSuchElementException();
  319           return x;
  320       }

注意：它移动头部而不是将所有元素向下移动到数组中。

于 2013-01-10T10:45:45.187 回答
16

尝试

    String[] a = {"1", "2"};
    String[] a2 = new String[a.length + 1];
    a2[0] = "0";
    System.arraycopy(a, 0, a2, 1, a.length);
于 2013-01-10T10:49:07.593 回答
14

如果你已经在使用 Guava，你可以ObjectArrays::concat这样做：

String[] args = ...;
ObjectArrays.concat("prepended", args);
于 2016-10-10T08:56:51.503 回答
6

这是@matteosilv 提出的解决方案的更正版本：

String[] myArray= {"hi","hi2"};
List<String> list = new LinkedList<String>(Arrays.asList(myArray));
list.add(0, "h3");
myArray = list.toArray(new String[list.size()]);
于 2014-01-09T20:07:13.930 回答
4

你不能......你必须向前移动它后面的所有字符串以适应新字符串。如果您直接将其添加到第 0 个索引，您将丢失那里的前一个元素

于 2013-01-10T10:44:13.250 回答
3 
String[] myArray= {"hi","hi2"};
List<String> temp = new ArrayList<String>(Arrays.asList(prova));
temp.add(0, "h3");
myArray = temp.toArray(new String[temp.size()]);
于 2013-01-10T11:02:22.443 回答
2

您可以使用流

  private Object[] prepend(String prefix, String[] row) {
        Stream<String> stream = Stream.concat(Arrays.asList(prefix).stream(), Arrays.asList(row).stream());
        return stream.toArray();
    }

打电话

prepend("one",new String[]{"two","three","four"})

结果

$1 ==> 对象[4] {“一”、“二”、“三”、“四”}

于 2021-08-20T03:49:07.120 回答
1

为此，您应该使用List.

如果您想特别使用内部数组，请使用ArrayList

于 2013-01-10T10:45:15.137 回答
1

我能做到的最好...

public static void main(String[] args) {
        String[] s = new String[] { "a", "b", "c" };
        System.out.println(Arrays.toString(prepend(s,"d")));
}

public static String[] prepend(String[] a, String el) {
        String[] c = new String[a.length+1];
        c[0] = el;
        System.arraycopy(a, 0, c, 1, a.length);
        return c;
}
于 2013-01-10T11:10:18.343 回答
1

您可以执行以下操作

public class Test {

public static String[] addFirst(String s[], String e) {
    String[] temp = new String[s.length + 1];
    temp[0] = e;
    System.arraycopy(s, 0, temp, 1, s.length);
    return temp;
}

public static void main(String[] args) {
    String[] s = { "b", "c" };
    s = addFirst(s, "a");
    System.out.println(Arrays.toString(s));
}
}
于 2013-01-10T11:21:20.997 回答