ArrayList<String>arraylist2=newArrayList<>();//1 - Remove an element from the specified index positionarraylist.remove(indexPosition);//2 - Remove the first occurence of element by its valuearraylist.remove(element);//3 - Remove all elements of the specified collection from arraylistarraylist.remo...
int length : 要copy的数组的长度 如果一个arraylist集合有0,1,2,3,4,5的数据 然后remove(3)从原数组的下标为4就是4开始,复制两个长度也就是4和5复制这两个, 接着 从目标数组开始(这里也是当前数组)的下标为3这里将其覆盖也就是变成0,1,2,4,5,5, 最后将最后一位置为null就变成0,1,2,4,5,null ...
}privateclassItrimplementsIterator<E> {intcursor;// index of next element to returnintlastRet=-1;// index of last element returned; -1 if no suchintexpectedModCount=modCount;//省略部分实现} Itr是ArrayList中的内部类,所以list.iterator()的作用是返回了一个Itr对象赋值到var2,后面调用var2.hasNext...
/** * An optimized version of AbstractList.Itr */ private class Itr implements Iterator<E> { int cursor; // index of next element to return int lastRet = -1; // index of last element returned; -1 if no such int expectedModCount = modCount; public boolean hasNext() { return cursor...
int lastRet = -1; // index of last element returned; -1 if no such int expectedModCount = modCount; //省略部分实现 } Itr是ArrayList中的内部类,所以list.iterator()的作用是返回了一个Itr对象赋值到var2,后面调用...
都说ArrayList在用foreach循环的时候,不能add元素,也不能remove元素,可能会抛异常,那我们就来分析一下它具体的实现。我目前的环境是java8。 有下面一段代码: public class TestForEachList extends BaseTests { @Test public void testForeach() {
ArrayList remove() removes the first occurrence of the specified element from this list, if it is present, else the list remains unchanged.
public E get(int index) { rangeCheck(index); return elementData(index); } public E set(int index, E element) { rangeCheck(index); E oldValue = elementData(index); elementData[index] = element; return oldValue; } 检查范围,设置,或者返回。 size,isEmpty,contains,indexOf,lastIndexOf:送分...
1. Removing an element from Array using for loop 2. Deleting an array element by its value 3. Deleting element by its value when the array contains duplicates 4. Shifting elements in the same array 5. Deleting elements from ArrayList Manual shifting of elements Using System.arraycopy() Conver...
Last but not least, we are traversing the set, which, in the worst scenario, will take O(N) time (if every element is unique) As a result, this method’s worst-case time complexity to remove duplicates from an array is O(N)+O(NlogN)+O(N)=O(NlogN). Auxiliary Space Analysis For...