Question

我正在尝试删除以r和g开头的列表元素

一个条件可以很好地删除从r开始的elememts，但是下一个在if循环中不起作用的条件可以帮助我弄错我的地方。谢谢.. !!

我的代码是

import java.util.Arrays;
import java.util.LinkedList;

public class EasyRemovingList {
    void removeList(){
    String[] inpArray={"red","green","blue","ivory"};
    LinkedList<String> upList = new LinkedList<String>(Arrays.asList(inpArray));
    System.out.println("Actual "+upList+" "+upList.size());
    //System.out.println(upList.get(3));
    for(int i = 0;i<upList.size();i++){
        //System.out.println(upList.get(i));    
    if(upList.get(i).startsWith("r")||upList.get(i).startsWith("g")){
        upList.remove(i);
        System.out.println("Updated "+upList);
    }
    }
    //System.out.println(upList);
    }
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        EasyRemovingList er = new EasyRemovingList();
        er.removeList();

    }

}

输出为

Actual [red, green, blue, ivory] 4
Updated [green, blue, ivory]

预期输出为

Actual [red, green, blue, ivory] 4
Updated [blue, ivory]

Answer 1

尝试使用removeIf()。

upList.removeIf(element -> {return element.starts("r") || element.startsWith("g");});

MVCE

import java.util.Arrays;
import java.util.LinkedList;

/**
 *
 * @author Sedrick
 */
public class JavaApplication4 {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        String[] inpArray={"red","green","blue","ivory"};
        LinkedList<String> upList = new LinkedList(Arrays.asList(inpArray));
        System.out.println("Actual "+upList+" "+upList.size());

        upList.removeIf(element -> {
            return element.startsWith("r") || element.startsWith("g"); //To change body of generated lambdas, choose Tools | Templates.
        });

        System.out.println("Actual "+upList+" "+upList.size());

    }

}

结果

run:
Actual [red, green, blue, ivory] 4
Actual [blue, ivory] 2
BUILD SUCCESSFUL (total time: 0 seconds)

Answer 2

您正在移动时正在移走，这在编码中是一种难闻的气味。

您可以尝试使用 iterator 进行删除。

通过iterator，您可以将其用作

while(iter.hasNext()) {
    String s = iter.next();
    if (s. startsWith(“r”) || s. startsWith(“g”)) {
          iter.remove();
    }
}

Answer 3

当您删除一个元素时，所有其他元素都将移过来，您将跳过下一个元素。为了避免这种情况，只需从1减去LinkedList的长度开始，然后循环直到i等于零：

for(int i = upList.size()-1;i>=0;i--){    
    if(upList.get(i).startsWith("r")||upList.get(i).startsWith("g")){
        upList.remove(i);        
    }
}
System.out.println("Updated "+upList);
}

输出：

Actual [red, green, blue, ivory] 4
Updated [blue, ivory]

Answer 4

您可以使用Java 8作为简单的解决方案：

import java.util.Arrays;
import java.util.LinkedList;
import java.util.function.Predicate;
import java.util.stream.Collectors;

public class StackOverFlow {

    public static void main(String[] args) {
        String[] inpArray = { "red", "green", "blue", "ivory" };
        LinkedList<String> upList = new LinkedList<String>(Arrays.asList(inpArray));
        System.out.println("Actual " + upList + " " + upList.size());

        Predicate<String> p = e -> e.startsWith("r") || e.startsWith("g");

        // Return a new list
        System.out.println(upList.stream().filter(p.negate()).collect(Collectors.toList()));

        // Updates your existing list
        upList.removeIf(p);
        System.out.println(upList);

    }

}

在Java中使用startsWith方法删除LinkedList元素

4 个答案: