嗨，我正在尝试制作删除方法。但是我不知道如何做到这一点。这是我的代码。
这是算法第四版中的LinkedList.java。
4单元测试
/**
* <i>Algorithms, 4th Edition</i> by Robert Sedgewick and Kevin
* Wayne.
* @author Robert Sedgewick
* @author Kevin Wayne
*/
public class LinkedQueue<Item> implements Iterable<Item> {
private int N; // number of elements on queue
private Node first; // beginning of queue
private Node last; // end of queue
// helper linked list class
private class Node {
private Item item;
private Node next;
}
/**
* Initializes an empty queue.
*/
public LinkedQueue() {
first = null;
last = null;
N = 0;
assert check();
}
/**
* Is this queue empty?
* @return true if this queue is empty; false otherwise
*/
public boolean isEmpty() {
return first == null;
}
/**
* Returns the number of items in this queue.
* @return the number of items in this queue
*/
public int size() {
return N;
}
/**
* Returns the item least recently added to this queue.
* @return the item least recently added to this queue
* @throws java.util.NoSuchElementException if this queue is empty
*/
public Item peek() {
if (isEmpty()) throw new NoSuchElementException("Queue underflow");
return first.item;
}
/**
* Adds the item to this queue.
* @param item the item to add
*/
public void enqueue(Item item) {
Node oldlast = last;
last = new Node();
last.item = item;
last.next = null;
if (isEmpty()) first = last;
else oldlast.next = last;
N++;
assert check();
}
/**
* Removes and returns the item on this queue that was least recently added.
* @return the item on this queue that was least recently added
* @throws java.util.NoSuchElementException if this queue is empty
*/
public Item dequeue() {
if (isEmpty()) throw new NoSuchElementException("Queue underflow");
Item item = first.item;
first = first.next;
N--;
if (isEmpty()) last = null; // to avoid loitering
assert check();
return item;
}
/**
* Returns a string representation of this queue.
* @return the sequence of items in FIFO order, separated by spaces
*/
public String toString() {
StringBuilder s = new StringBuilder();
for (Item item : this)
s.append(item + " ");
return s.toString();
}
// check internal invariants
private boolean check() {
if (N == 0) {
if (first != null) return false;
if (last != null) return false;
}
else if (N == 1) {
if (first == null || last == null) return false;
if (first != last) return false;
if (first.next != null) return false;
}
else {
if (first == last) return false;
if (first.next == null) return false;
if (last.next != null) return false;
// check internal consistency of instance variable N
int numberOfNodes = 0;
for (Node x = first; x != null; x = x.next) {
numberOfNodes++;
}
if (numberOfNodes != N) return false;
// check internal consistency of instance variable last
Node lastNode = first;
while (lastNode.next != null) {
lastNode = lastNode.next;
}
if (last != lastNode) return false;
}
return true;
}
//working properly
void reverseBystack(){
Stack<Item> s = new Stack<>();
Item item;
while (isEmpty() != true){
item = dequeue();
s.push(item);
}
while(s.isEmpty() != true){
item = s.pop();
enqueue(item);
}
}
//working properly.
void reverseBylink() {
Node prev = null;
Node current = this.first;
Node next = null;
Node temp = null;
while (current != null) {
next = current.next;
current.next = prev;
prev = current;
current = next;
}
temp =first;
first = last;
last = temp;
}
//How to do this...;<..
int remove(Item item) {
Node cur = this.first;
while (cur !=null) {
if (cur.item.equals(item)) {
item = dequeue();
}
cur = cur.next;
N++;
}
return 0;
}
/**
* Returns an iterator that iterates over the items in this queue in FIFO order.
* @return an iterator that iterates over the items in this queue in FIFO order
*/
public Iterator<Item> iterator() {
return new ListIterator();
}
// an iterator, doesn't implement remove() since it's optional
private class ListIterator implements Iterator<Item> {
private Node 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;
}
}
输出。
/**
* Unit tests the <tt>LinkedQueue</tt> data type.
*/
public static void main(String[] args) {
LinkedQueue<String> q = new LinkedQueue<String>();
/* Working properly for reverseByStack.
q.enqueue("a");
q.enqueue("b");
q.enqueue("c");
q.enqueue("a");
q.enqueue("b");
q.enqueue("d");
q.enqueue("b");
q.enqueue("abba");
q.enqueue("a");
q.enqueue("z");
q.enqueue("a");
q.reverseBystack();
System.out.println(q);
StdOut.println("(" + q.size() + " left on queue)");
*/
/*Move on to next, working properly
q.enqueue("a");
q.enqueue("b");
q.enqueue("c");
q.enqueue("a");
q.enqueue("b");
q.enqueue("d");
q.enqueue("b");
q.enqueue("abba");
q.enqueue("a");
q.enqueue("z");
q.enqueue("a");
q.reverseBylink();
System.out.println(q);
StdOut.println("(" + q.size() + "left on queue)");*/
q.enqueue("a");
q.enqueue("b");
q.enqueue("c");
q.enqueue("a");
q.enqueue("b");
q.enqueue("d");
q.enqueue("b");
q.enqueue("abba");
q.enqueue("a");
q.enqueue("z");
q.enqueue("a");
System.out.println(q);
System.out.println("Remove some of elements. and use reverseByLink");
q.remove("a");
q.remove("f");
q.remove("c");
System.out.println(q);
}
删除一些元素。并使用reverseByLink
a b c a b d b abba a z a
我不知道为什么在abba之后不删除String a。
答案 0 :(得分：1)
_renderItem (item) {
return(
<View style={{ width: 350, flexGrow: 1, }}>
<Text style={{ fontSize: 16, color: 'black', }}>
{item.law_practice_description} , // item.law_practice_description shows me the text like Administrative Adjudications etc
</Text>
</View>
);
}
render() {
return (
<View style={{ flex: 1 }}>
<Text style={styles.titleTxt}>Administrative Law</Text>
<FlatList
style={{ marginTop: 20,}}
data={this.state.data}
renderItem={({item}) => this._renderItem(item) }
keyExtractor={(index) => index.toString()}
/>
</View>
);
}
}
答案 1 :(得分：0)
我相当确定此方法是错误的：
int remove(Item item) {
Node cur = this.first;
while (cur !=null) {
if (cur.item.equals(item)) {
item = dequeue();
}
cur = cur.next;
N++;
}
return 0;
}
您的dequeue方法将弹出列表的顶部。但是您要删除的项目可能不在列表的前面。
我没有寻找更多的问题。
这是标准的链接列表内容。您的某些教科书应该对此具有算法。但基本上，您需要跟踪使用的最后一个指针。假设您有lastPtr和currentPtr，并且确定需要currentPtr。
然后lastPtr.next = currentPtr.next
如果您高枕无忧，那就更有趣了。您需要认识到这一点，而要做first.next = currentPtr.next。
答案 2 :(得分：0)
dequeue（）方法从前面弹出项目，但是要删除所有出现的任何字符串，您需要将remove（）方法修改为-
.eml