用java源代码学数据结构<四>: LinkedList 详解(三)
d = newNode;
}
if (succ == null) {
last = pred;
} else {
//将保存的值恢复
pred.next = succ;
succ.prev = pred;
}
size += numNew;
modCount++;
return true;
}
public void clear() {
// Clearing all of the links between nodes is "unnecessary", but:
// - helps a generational GC if the discarded nodes inhabit
// more than one generation
// - is sure to free memory even if there is a reachable Iterator
for (Node x = first; x != null; ) {
Node next = x.next;
//将节点设置为全null,然后由gc回收
x.item = null;
x.next = null;
x.prev = null;
x = next;
}
first = last = null;
size = 0;
modCount++;
}
public E get(int index) {
checkElementIndex(index);
return node(index).item;
}
//将此列表中指定位置的元素替换为指定的元素。返回旧的元素
public E set(int index, E element) {
checkElementIndex(index);
Node x = node(index);
E oldVal = x.item;
x.item = element;
return oldVal;
}
//在此列表中指定的位置插入指定的元素。
public void add(int index, E element) {
checkPositionIndex(index);
if (index == size)
linkLast(element);
else
linkBefore(element, node(index));
}
public E remove(int index) {
checkElementIndex(index);
return unlink(node(index));
}
//元素有效位置参数判断
private boolean isElementIndex(int index) {
return index >= 0 && index < size;
}
//index==size，表示从结尾开始
private boolean isPositionIndex(int index) {
return index >= 0 && index <= size;
}
private String outOfBoundsMsg(int index) {
return "Index: "+index+", Size: "+size;
}
private void checkElementIndex(int index) {
if (!isElementIndex(index))
throw new IndexOutOfBoundsException(outOfBoundsMsg(index));
}
private void checkPositionIndex(int index) {
if (!isPositionIndex(index))
throw new IndexOutOfBoundsException(outOfBoundsMsg(index));
}
//返回指定位置的node对象
Node node(int index) {
// assert isElementIndex(index);
//看index离头结点近，还是离尾节点近，提高效率
if (index < (size >> 1)) {
Node x = first;
for (int i = 0; i < index; i++)
x = x.next;
return x;
} else {
Node x = last;
for (int i = size - 1; i > index; i--)
x = x.prev;
return x;
}
}
public int indexOf(Object o) {
//对o为null或其他对象，使用==或equals方法
int index = 0;
if (o == null) {
for (Node x = first; x != null; x = x.next) {
if (x.item == null)
return index;
index++;
}
} else {
for (Node x = first; x != null; x = x.next) {
if (o.equals(x.item))
return index;
index++;
}
}
return -1;
}
//反向查找
public int lastIndexOf(Object o) {
int index = size;
if (o == null) {
for (Node x = last; x != null; x = x.prev) {
index--;
if (x.item == null)