Java - List遍曆、判斷、刪除元素時的陷阱

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開發中,常有“遍曆集合,依次判斷是否符合條件,如符合條件則刪除當前元素”的情境,有一些陷阱常犯。

 

漏網之魚
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);    }}

 

日誌列印:

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

如日誌所見,其中值為4的元素並未經過判斷,漏網之魚。

 

對於此情況,我一般都從後面開始遍曆,以避免問題:

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);    }}

 

或直接新開一個list,重新擺放,但浪費記憶體,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異常

用Iterator方式或簡寫的for(Object o : list) {}方式,遍曆集合,修改元素時會報異常,具體見關於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);    }}

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);    }}

 

日誌:

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

在刪除元素“3”時,會報異常。

 

對於此情況,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);    }}

 

Java - List遍曆、判斷、刪除元素時的陷阱

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