正确使用ArrayList的sublist
1.问题描述
使用subList方法获取的list,只是一个视图,而不是一个新的list,如果进行操作,会导致原list也变化了。
public class Test {
public static void main(String[] args){
List<Integer> list=new ArrayList<>();
for(int i=0;i<100;i++){
list.add(i);
}
for(int i=0;i<10;i++){
List<Integer> temp=list.subList(0,10);
System.out.println(temp);
Iterator<Integer> iterator = temp.iterator();
while(iterator.hasNext()){
iterator.next();
iterator.remove();
}
}
System.out.println(list);
}
}
打印结果:
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
[10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
[20, 21, 22, 23, 24, 25, 26, 27, 28, 29]
[30, 31, 32, 33, 34, 35, 36, 37, 38, 39]
[40, 41, 42, 43, 44, 45, 46, 47, 48, 49]
[50, 51, 52, 53, 54, 55, 56, 57, 58, 59]
[60, 61, 62, 63, 64, 65, 66, 67, 68, 69]
[70, 71, 72, 73, 74, 75, 76, 77, 78, 79]
[80, 81, 82, 83, 84, 85, 86, 87, 88, 89]
[90, 91, 92, 93, 94, 95, 96, 97, 98, 99]
[]
2.源码分析
ArrayList的sublist方法并没有新建一个ArrayList,而是创建了一个原ArrayList的视图。
public List<E> subList(int fromIndex, int toIndex) {
subListRangeCheck(fromIndex, toIndex, size);
return new SubList(this, 0, fromIndex, toIndex);
}
这个视图,是一个ArrayList内部类,其实就是用一些坐标指向原ArrayList,所以如果对这个SubList进行操作,也会影响到原ArrayList。
private class SubList extends AbstractList<E> implements RandomAccess {
private final AbstractList<E> parent;
private final int parentOffset;
private final int offset;
int size;
SubList(AbstractList<E> parent,
int offset, int fromIndex, int toIndex) {
this.parent = parent;
this.parentOffset = fromIndex;
this.offset = offset + fromIndex;
this.size = toIndex - fromIndex;
this.modCount = ArrayList.this.modCount;
}
......
}
3.解决
可以自己新建一个ArrayList,这样就不会导致上面的问题了。
public class Test {
public static void main(String[] args){
List<Integer> list=new ArrayList<>();
for(int i=0;i<100;i++){
list.add(i);
}
for(int i=0;i<10;i++){
List<Integer> temp=new ArrayList<>(list.subList(10*i,10*(i+1)));
System.out.println(temp);
Iterator<Integer> iterator = temp.iterator();
while(iterator.hasNext()){
iterator.next();
iterator.remove();
}
}
System.out.println(list);
}
}
打印结果
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
[10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
[20, 21, 22, 23, 24, 25, 26, 27, 28, 29]
[30, 31, 32, 33, 34, 35, 36, 37, 38, 39]
[40, 41, 42, 43, 44, 45, 46, 47, 48, 49]
[50, 51, 52, 53, 54, 55, 56, 57, 58, 59]
[60, 61, 62, 63, 64, 65, 66, 67, 68, 69]
[70, 71, 72, 73, 74, 75, 76, 77, 78, 79]
[80, 81, 82, 83, 84, 85, 86, 87, 88, 89]
[90, 91, 92, 93, 94, 95, 96, 97, 98, 99]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99]
本文解析了Java中ArrayList的sublist方法原理,指出其返回的是原列表的视图而非副本,直接修改会导致原列表变化。通过源码分析,提供了解决方案,即创建新的ArrayList来避免影响原列表。
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