Java LinkedList删除零长度成员

Question

我试图从长度为零的LinkedList中删除成员。我试图使代码动态，它从字符串中删除所有空格，但留下的单词分开和顺序。我尝试通过将字符串转换为数组，并在空格处分割它来实现这一点。然后我根据自己的喜好将其转换为LinkedList。

private void sliceItUpFreashness() {
        String s = "    hello there"; // four spaces at beginning of string
        String[] sa = s.split(" ");
        LinkedList<String> ll = new LinkedList<>(Arrays.asList(sa));
        for (int i = 0; i < ll.size(); i++) {
            System.out.println("Size for ll("+i+") == "+ll.get(i).length()); // confirms length of the members is 0
            if (ll.get(i).length() == 0) {
                ll.remove(i);
            }
        }
        System.out.println("-----");
        for (String a : ll) { // this confirms if the zero-length members were removed
            System.out.println(a);
        }
    }

但是上面的代码并没有删除长度为0或null的成员。我能做什么?

Answer 1

我建议您迭代一次数组并直接构建LinkedList -

String s = "    hello there"; // four spaces at beginning of string
String[] sa = s.split(" ");
List<String> ll = new LinkedList<>();
for (String str : sa) {
  if (str.length() > 0) { // <-- check for empty string.
    ll.add(str);
  }
}

另一种解决方案是迭代String本身并跳过split，

List<String> ll = new LinkedList<>();
StringBuilder sb = new StringBuilder();
for (char ch : s.toCharArray()) { // <-- String toCharArray
  if (ch != ' ') {
    sb.append(ch); // <-- not a space.
  } else {
    if (sb.length() > 0) { // <-- add non zero length strings to List
      ll.add(sb.toString());
      sb.setLength(0);
    }
  }
}
// Add the last of the buffer
if (sb.length() > 0) {
  ll.add(sb.toString());
}

Answer 2

或者使用Collections.singleton("")

    String s = "    hello there"; // four spaces at beginning of string
    String[] sa = s.split(" ");
    List<String> list = new ArrayList<String>(Arrays.asList(sa));
    list.removeAll(Collections.singleton(""));

Answer 3

代码:

String s = "    hello there"; // four spaces at beginning of string
    String[] sa = s.split(" ");
    LinkedList<String> ll = new LinkedList<>(Arrays.asList(sa));
    ll.stream()
      .filter( s1 -> !s1.isEmpty())
      .map(s1-> s1)
      .forEach(System.out::print);

输出:

    hello there
hellothere

希望有帮助，如果你需要更多的解释，请告诉我