java - 关于单链表实现的一个stack的问题
背景：菜鸟一枚，想补补基础。最近在看算法
，1.3.26有一道题是：编写一个方法remove()接受一条链表和一个字符串key作为参数，删除链表中所有的item域为key的节点。
上代码：
public class Stack<Item> implements Iterable<Item> {
private Node<Item> first; // top of stack
private int n; // size of the stack
// helper linked list class
private static class Node<Item> {
private Item item;
private Node<Item> next;
}
/**
* Initializes an empty stack.
*/
public Stack() {
first = null;
n = 0;
}
/**
* Returns true if this stack is empty.
*
* @return true if this stack is empty; false otherwise
*/
public boolean isEmpty() {
return first == null;
}
/**
* Returns the number of items in this stack.
*
* @return the number of items in this stack
*/
public int size() {
return n;
}
/**
* Adds the item to this stack.
*
* @param item
* the item to add
*/
public void push(Item item) {
Node<Item> oldfirst = first;
first = new Node<Item>();
first.item = item;
first.next = oldfirst;
n++;
}
/**
* Removes and returns the item most recently added to this stack.
*
* @return the item most recently added
* @throws NoSuchElementException
* if this stack is empty
*/
public Item pop() {
if (isEmpty())
throw new NoSuchElementException("Stack underflow");
Item item = first.item; // save item to return
first = first.next; // delete first node
n--;
return item; // return the saved item
}
/**
* Returns (but does not remove) the item most recently added to this stack.
*
* @return the item most recently added to this stack
* @throws NoSuchElementException
* if this stack is empty
*/
public Item peek() {
if (isEmpty())
throw new NoSuchElementException("Stack underflow");
return first.item;
}
/**
* Returns a string representation of this stack.
*
* @return the sequence of items in this stack in LIFO order, separated by
* spaces
*/
public String toString() {
StringBuilder s = new StringBuilder();
for (Item item : this)
s.append(item + " ");
return s.toString();
}
// 1.3.20
public void delete(int k) {
// 不允许大于n的值
if (k <= n) {
Node<Item> nowNextNode = first;
// 要第几个就next到第几个的上一个
for (int i = 1; (k - 1) != i; i++) {
nowNextNode = nowNextNode.next;
}
// 删除节点
Node<Item> needLinePoint = nowNextNode.next.next;
nowNextNode.next = needLinePoint;
} else {
// 这里简单的打印一下
StdOut.print("!------------------删除元素越界！------------------!");
}
}
// 1.3.21
public boolean find(String key) {
// 其实也可以用foreach
Node<Item> nowNode = first;
while (true) {
if (nowNode.item.equals(key)) {
StdOut.print("output : true");
return true;
// 链表没有下一个了
} else if (first.next.equals(null)) {
StdOut.print("output : false");
return false;
}
nowNode = nowNode.next;
StdOut.println("then next~");
}
}
// 1.3.24
public void removeAfter(Node<Item> node) {
if (node == null || node.next == null) {
// 什么也别做
} else {
if (isEmpty())
throw new NoSuchElementException("Stack underflow");
Node<Item> tempNode = node.next.next;
node.next = null;
node.next = node.next.next;
n--;
}
}
//1.3.26 接受一个链表和一个字符串key作为参数，删除链表中所有item域为key的节点
// 那么为了降低难度，直接写成对this产生作用的方法
public void remove(String key) {
//首先肯定是一个能够不停next直到没有的循环
//第二个就是要时时刻刻记住上一个节点，这样便于移除现在的节点
//然后就是对等于key和不等于key的逻辑判断
//先判断nowNode.next == null，如果是，说明已经到了尾部，结束循环
//相等：移除节点，上一个节点还是原先的上一个节点，但会连接至“被移除节点”的next，继续循环
//不相等：frontNode变更为nowNode，继续循环
Node<Item> nowNode = first;
Node<Item> frontNode = null;
while(true){
if (nowNode.next.equals(null)){
StdOut.print("没有拥有该字符串的item");
break;
}else if (nowNode.item.equals(key)){
frontNode.next = nowNode.next;
nowNode.next=null;
nowNode = frontNode.next;
StdOut.print("已搜索到该字符串！");
}else {
frontNode = nowNode;
nowNode = nowNode.next;
}
}
}
/**
* Returns an iterator to this stack that iterates through the items in LIFO
* order.
*
* @return an iterator to this stack that iterates through the items in LIFO
* order
*/
public Iterator<Item> iterator() {
return new ListIterator<Item>(first);
}
// an iterator, doesn't implement remove() since it's optional
private class ListIterator<Item> implements Iterator<Item> {
private Node<Item> current;
public ListIterator(Node<Item> first) {
current = first;
}
public boolean hasNext() {
return current != null;
}
public void remove() {
throw new UnsupportedOperationException();
}
public Item next() {
if (!hasNext())
throw new NoSuchElementException();
Item item = current.item;
current = current.next;
return item;
}
}
}
但是一经测试就报java.lang.NullPointerException，为什么？
详细错误：
java.lang.NullPointerException
at Unit1.Stack.remove(Stack.java:168)
at Unit1.StackTest.Test(StackTest.java:19)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(Unknown Source)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(Unknown Source)
at java.lang.reflect.Method.invoke(Unknown Source)
at org.junit.runners.model.FrameworkMethod$1.runReflectiveCall(FrameworkMethod.java:50)
at org.junit.internal.runners.model.ReflectiveCallable.run(ReflectiveCallable.java:12)
at org.junit.runners.model.FrameworkMethod.invokeExplosively(FrameworkMethod.java:47)
at org.junit.internal.runners.statements.InvokeMethod.evaluate(InvokeMethod.java:17)
at org.junit.runners.ParentRunner.runLeaf(ParentRunner.java:325)
at org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:78)
at org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:57)
at org.junit.runners.ParentRunner$3.run(ParentRunner.java:290)
at org.junit.runners.ParentRunner$1.schedule(ParentRunner.java:71)
at org.junit.runners.ParentRunner.runChildren(ParentRunner.java:288)
at org.junit.runners.ParentRunner.access$000(ParentRunner.java:58)
at org.junit.runners.ParentRunner$2.evaluate(ParentRunner.java:268)
at org.junit.runners.ParentRunner.run(ParentRunner.java:363)
at org.eclipse.jdt.internal.junit4.runner.JUnit4TestReference.run(JUnit4TestReference.java:86)
at org.eclipse.jdt.internal.junit.runner.TestExecution.run(TestExecution.java:38)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.runTests(RemoteTestRunner.java:459)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.runTests(RemoteTestRunner.java:678)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.run(RemoteTestRunner.java:382)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.main(RemoteTestRunner.java:192)
测试里相关的代码：
public class StackTest {
@Test
public void Test() {
Stack<String> stack = new Stack<String>();
while (!StdIn.isEmpty()) {
String item = StdIn.readString();
if (item.equals("find")) {
stack.find("3");
}
else if (item.equals("remove")) {
stack.remove("1");
}
else if (item.equals("del")) {
stack.delete(3);
}
else if (!item.equals("-")) {
stack.push(item);
}
else if (!stack.isEmpty()) {
StdOut.print(stack.pop() + " ");
}
}
StdOut.println("(" + stack.size() + " left on stack)");
}
}报错行是stack里的
if (nowNode.next.equals(null)){以及测试里的
stack.remove("1");但是我认为一楼的答案并不是很好...找到空指针的引用后，就向前找这个引用的初始化的逻辑部分。
初始化部分：
Node<Item> nowNode = first;
Node<Item> frontNode = null;我很同意授人鱼不如授人以渔。然而...
StdIn是一个能够不停输入的函数。
测试数据输入的是
1
2
3
4
5
remove
感谢各位！问题似乎已经解决了！附上代码：
public void remove(String key) {
// 首先肯定是一个能够不停next直到没有的循环
// 然后就是对等于key和不等于key的逻辑判断
// 先判断nowNode== null，如果是，说明已经到了尾部，结束循环。同样nowNode.next == null也是
Node<Item> nowNode = first;
while (true) {
if (nowNode == null ) {
StdOut.println("搜索结束");
break;
}
if (nowNode.item.equals(key)) {
StdOut.println("已搜索到该字符串！");
if (nowNode.next == null){
StdOut.println("已到达底端，退出！");
break;
}
//删除节点操作
//临时存储的对象：直接让nextNode和nowNode交换
Node<Item> nextNode ;
nextNode = nowNode.next;
nowNode.item = nextNode.item;
nowNode.next = nextNode.next;
nextNode = null;
}
else {
nowNode = nowNode.next;
}
}
}
