Java - List遍历、判断、删除元素时的陷阱

开发中，常有“遍历集合，依次判断是否符合条件，如符合条件则删除当前元素”的场景，有一些陷阱常犯。

漏网之鱼

import java.util.ArrayList;
import java.util.List;

public class ListTest_Unwork {

    public static void main(String[] args) {
        List<String> list = new ArrayList<String>();
        list.add("1");
        list.add("2");
        list.add("3");
        list.add("4");
        list.add("5");
        System.out.println("Original list : " + list);

        String temp = null;
        for (int i = 0; i < list.size(); i++) {
            temp = list.get(i);

            System.out.println("Check for " + temp);
            if ("3".equals(temp)) {
                list.remove(temp);
            }
        }
        System.out.println("Removed  list : " + list);
    }

}

日志打印：

Original list : [1, 2, 3, 4, 5]
Check for 1
Check for 2
Check for 3
Check for 5
Removed  list : [1, 2, 4, 5]

如日志所见，其中值为4的元素并未经过判断，漏网之鱼。

对于此情况，我一般都从后面开始遍历，以避免问题：

import java.util.ArrayList;
import java.util.List;

public class ListTest_Work {

    public static void main(String[] args) {
        List<String> list = new ArrayList<String>();
        list.add("1");
        list.add("2");
        list.add("3");
        list.add("4");
        list.add("5");
        System.out.println("Original list : " + list);
        System.out.println();

        String temp = null;
        for (int i = list.size() - 1; i >= 0; i--) {
            temp = list.get(i);

            System.out.println("Check for " + temp);
            if ("3".equals(temp)) {
                list.remove(temp);
            }
        }
        System.out.println("Removed  list : " + list);
    }

}

或直接新开一个list，重新摆放，但浪费内存，慎用：

import java.util.ArrayList;
import java.util.List;

public class ListTest_Work2 {

    public static void main(String[] args) {
        List<String> list = new ArrayList<String>();
        list.add("1");
        list.add("2");
        list.add("3");
        list.add("4");
        list.add("5");
        System.out.println("Original list : " + list);
        System.out.println();

        List<String> tempList = new ArrayList<String>();
        for (String temp : list) {
            System.out.println("Check for " + temp);
            if (!"3".equals(temp)) {
                tempList.add(temp);
            }
        }
        System.out.println("Removed  list : " + tempList);
    }

}

ConcurrentModificationException异常

用Iterator方式或简写的for(Object o : list) {}方式，遍历集合，修改元素时会报异常，具体见关于List的ConcurrentModificationException

import java.util.ArrayList;
import java.util.List;

public class ListTest2_Unwork {

    public static void main(String[] args) {
        List<String> list = new ArrayList<String>();
        list.add("1");
        list.add("2");
        list.add("3");
        list.add("4");
        list.add("5");
        System.out.println("Original list : " + list);
        System.out.println();

        for (String temp : list) {
            System.out.println("Check for " + temp);
            if ("3".equals(temp)) {
                list.remove(temp);
            }
        }
        System.out.println("Removed  list : " + list);
    }

}

或

import java.util.ArrayList;
import java.util.Iterator;
import java.util.List;

public class ListTest3_Unwork {

    public static void main(String[] args) {
        List<String> list = new ArrayList<String>();
        list.add("1");
        list.add("2");
        list.add("3");
        list.add("4");
        list.add("5");
        System.out.println("Original list : " + list);
        System.out.println();

        Iterator<String> i = list.iterator();
        String temp = null;
        while (i.hasNext()) {
            temp = i.next();
            System.out.println("Check for " + temp);
            if ("3".equals(temp)) {
                list.remove(temp);
            }
        }
        System.out.println("Removed  list : " + list);
    }

}

日志：

Original list : [1, 2, 3, 4, 5]

Check for 1
Check for 2
Check for 3
Exception in thread "main" java.util.ConcurrentModificationException
    at java.util.ArrayList$Itr.checkForComodification(ArrayList.java:859)
    at java.util.ArrayList$Itr.next(ArrayList.java:831)
    at ListTest3_Unwork.main(ListTest3_Unwork.java:20)

在删除元素“3”时，会报异常。

对于此情况，需要用iterator的remove方法替代：

import java.util.ArrayList;
import java.util.Iterator;
import java.util.List;

public class ListTest3_Work {

    public static void main(String[] args) {
        List<String> list = new ArrayList<String>();
        list.add("1");
        list.add("2");
        list.add("3");
        list.add("4");
        list.add("5");
        System.out.println("Original list : " + list);
        System.out.println();

        Iterator<String> i = list.iterator();
        String temp = null;
        while (i.hasNext()) {
            temp = i.next();
            System.out.println("Check for " + temp);
            if ("3".equals(temp)) {
                i.remove();
            }
        }
        System.out.println("Removed  list : " + list);
    }

}