如何 return 字符串不多余

Question

我有一个用于链表的 toString 方法，它像时尚一样以数组形式打印出来，但它也会打印出多余的字符。我如何摆脱最后一个 ", "?

public String toString() {
  String ret = "[";
  Node current = head;
  while(current.getNext() != null) {
    current = current.getNext();
    ret += current.get() + ", ";
  }
  return ret + "]";
}

示例：我们有一个包含 3 个元素的链表，然后我们将其打印出来。

"[1, 2, 3, ]"

Answer 1

我没有测试过，但应该可以。

public String toString() {
  String ret = "[";
  Node current = head;
  while(current.getNext() != null) {
    current = current.getNext();
    ret += current.get();
    if(current.getNext() != null) ret += ", ";
  }
  return ret + "]";
}

Answer 2

您可以使用 substring 删除最后 2 个字符。

return ret.substring(0, ret.length()-2)+"]"

Answer 3

public String toString() {
  StringBuilder ret = new StringBuilder('[');
  Node current = head;
  while(current.getNext() != null) {
    current = current.getNext();
    ret.append(current.get()).append(", ");
  }
  if (ret.length() > 1) {
      ret.delete(ret.length() - 2, ret.length());
  }
  return ret.append(']').toString();
}

必须警惕一个空列表，所以它仍然是 []。这留下了两种可能性：布尔标志或之后删除。为了获得更好的性能，请使用 StringBuilder。在某种程度上，编译器会将字符串连接转换为 StringBuilder，但这里并非完全如此。

Answer 4

使用这个ret.substring(0, ret.length() - 1)

@Override
public String toString() {
      String ret = "[";
      Node current = head;
      while(current.getNext() != null) {
        current = current.getNext();
        ret += current.get() + ", ";
      }

      ret = ret.substring(0, ret.length() - 1);

      return ret + "]";
}

Answer 5

也许您应该考虑看一下 toString() Java 方法 AbstractCollection。这将使您了解如何为此目的使用 StringBuilder：

  /**
     * Returns a string representation of this collection.  The string
     * representation consists of a list of the collection's elements in the
     * order they are returned by its iterator, enclosed in square brackets
     * (<tt>"[]"</tt>).  Adjacent elements are separated by the characters
     * <tt>", "</tt> (comma and space).  Elements are converted to strings as
     * by {@link String#valueOf(Object)}.
     *
     * @return a string representation of this collection
     */
    public String toString() {
        Iterator<E> it = iterator();
        if (! it.hasNext())
            return "[]";

        StringBuilder sb = new StringBuilder();
        sb.append('[');
        for (;;) {
            E e = it.next();
            sb.append(e == this ? "(this Collection)" : e);
            if (! it.hasNext())
                return sb.append(']').toString();
            sb.append(',').append(' ');
        }
    }

Answer 6

试试这个：

public String toString() {
  String ret = "[";
  Node current = head;
  while(current.getNext() != null) {
    current = current.getNext();
    ret += current.get() + ", ";
  }

  ret=ret.substring(0,ret.length()-2);
  return ret + "]";
}

Answer 7

我使用Apache Commons Lang - StringUtils和ArrayList到return字符串没有多余的。看我怎么用：

public String toString() {
    List<String> ret = new ArrayList<String>();
    Node current = head;
    while (current.getNext() != null) {
        current = current.getNext();
        ret.add(current.get());
    }

    return "[" + StringUtils.join(ret, ",") + "]";
}

如何 return 字符串不多余

How to return string without excess

java

string

tostring