我目前正在进行Junit测试,以便在java中实现Doubly Linked列表。我理解它需要一遍又一遍地处理绘制的图片,但我不能让我的removeFirst方法工作。 addLast();
工作直到removeFirst被调用。
private Link<I> first;
private Link<I> last;
public boolean isEmpty() {
return size() == 0;
}
public int size() {
int count = 0;
Link<I> aLink = first;
while (aLink != null) {
count++;
aLink = aLink.getAfter();
}
return count;
}
public I get(int index) {
Link<I> aLink = first;
int count = 0;
while (count != index) {
aLink = aLink.getAfter();
count++;
}
return aLink.getItem();
}
public void addLast(I anItem) {
Link<I> aLink = new Link<I>(anItem);
if (isEmpty()) {
first = aLink;
first = last;
} else {
last.setAfter(aLink);
aLink.setBefore(last);
last = aLink;
}
}
public void addFirst(I anItem) {
Link<I> aLink = new Link<I>(anItem);
if (isEmpty()) {
first = aLink;
last = first;
} else {
aLink.setAfter(first);
first.setBefore(last);
first = aLink;
}
}
public I removeFirst() {
I removed = first.get(0);
if(size()==3) {
first = first.getAfter();
first.setBefore(null);
first.setAfter(last);
return removed;
} else if(size()==2) {
first = first.getAfter();
first.setBefore(null);
first.setAfter(null);
return removed;
}else {
first = null;
first.setBefore(null);
first.setAfter(null);
return removed;
}
}
public I removeLast() {
I removed = last.getItem();
if (isEmpty()) {
removed = null;
return removed;
} else {
last = last.getBefore();
removed = last.getItem();
return removed;
}
}
}
public class Link<I> {
private Link<I> after;
private Link<I> before;
private I item;
public Link(I anItem) {
item = anItem;
}
public Link<I> getAfter(){
return after;
}
public void setAfter(Link<I> aLink) {
after = aLink;
}
public Link<I> getBefore(){
return before;
}
public void setBefore(Link<I> aLink) {
before = aLink;
}
public I getItem() {
return item;
}
public void setItem(I anItem) {
item = anItem;
}
}
这是测试单元
void testAddLast() {
notes.addLast("do");
notes.addLast("re");
notes.addLast("mi");
String note = notes.removeFirst();
assertTrue(notes.size()==2);
assertTrue("do".equals(note));
note = notes.removeFirst();
assertTrue(notes.size()==1);
assertTrue("re".equals(note));
note= notes.removeFirst();
assertTrue(notes.isEmpty());
assertTrue("mi".equals(note));
note = notes.removeFirst();
assertTrue(note == null);
assertTrue(notes.isEmpty());
assertTrue(notes.size() == 0);
}
我觉得我现在拥有的是让我接近,但I removed = first.get(0);
不断给我一个空指针异常。两个多星期前,我在这项任务中获得了额外的时间,而且我仍然非常努力。我在调试器之后尝试了一堆removeFirst()方法的变体来帮助。
任何见解都会对我有所帮助。我有更多的测试用例可以添加到帖子中。
这似乎是给你带来麻烦的方法
public I removeFirst() {
I removed = first.get(0);
if(size()==3) {
first = first.getAfter();
first.setBefore(null);
first.setAfter(last);
return removed;
} else if(size()==2) {
first = first.getAfter();
first.setBefore(null);
first.setAfter(null);
return removed;
}else {
first = null;
first.setBefore(null);
first.setAfter(null);
return removed;
}
}
你说当你调用first.get(0)
时你得到一个NullPointerException。这意味着first
为空。
这是添加节点的方式:
public void addLast(I anItem) {
Link<I> aLink = new Link<I>(anItem);
if (isEmpty()) {
first = aLink;
first = last;
} else {
last.setAfter(aLink);
aLink.setBefore(last);
last = aLink;
}
}
您分配给first
aLink
,然后用last
覆盖该赋值,该值为null。
你可能意味着在那里说last = first
。
试试这个测试用例:
void testAddOne() {
assertTrue(notes.count() == 0);
notes.addLast("do");
assertTrue(notes.count() == 1);
}
通常,您的测试用例非常复杂。您希望更简单的测试来检查操作的后置条件。例如,插入一个元素应该将计数增加一。删除元素应该将计数减少一个。添加两个元素,check get返回第一个索引0,第二个返回索引1,依此类推。