java如何保持堆栈的最大值？
我的头撞在墙上已经一个星期了，似乎哪儿也去不了。我希望能够从堆栈中获取最大值，并将其保留在堆栈中，因此最终堆栈中只有一个值，即最大值。我相信我的保留最大值的算法是正确的，但我认为我的pop方法出现故障，我无法找出原因。感谢您的指导。我正在处理的两个文件都包括在内
import java.io.File;
import java.io.IOException;
import java.util.ListIterator;
import java.util.Random;
public class LinkedListStack {
DoublyLinkedList<Integer> list = new DoublyLinkedList<Integer>();
public int size() {
if (list.isEmpty()) {
System.out.println("Stack is empty");
}
return list.size();
}
public void push(Integer s) {
list.add(s);
}
public Integer pop() {
/* need help with this method */
}
public Integer top() {
return list.iterator().next();
}
public boolean isEmpty() {
return list.isEmpty();
}
public String displayStack() {
return (list.toString());
}
public static void main(String[] args) throws IOException {
LinkedListStack stack = new LinkedListStack();
int n, seed;
File outputFile;
File dir = new File(".");
if (args.length > 0) {
n = Integer.parseInt(args[0]);
seed = Integer.parseInt(args[1]);
}else {
n = 10;
seed = 1;
}
if (args.length == 3) {
outputFile = new File(args[2]);
} else {
outputFile = new File(dir.getCanonicalPath() + File.separator + "Files/testOut_Stack");
}
OutputWriter out = new OutputWriter(outputFile);
Random r = new Random(seed);
Integer nextval;
for (int i = 0; i < n; i++) {
nextval = r.nextInt(10000);
if (stack.isEmpty()) {
stack.push(nextval);
}
if (nextval > stack.top()) {
stack.pop();
stack.push(nextval);
}
/* retain the max value that you see among the integers generated in the stack.
* In the end there should be only one integer in stack, which is the max value
*/
}
// write the content of stack -- which is the max value -- to file
out.writeOutput(stack.displayStack());
}
}
import java.io.File;
import java.io.IOException;
import java.util.ListIterator;
import java.util.NoSuchElementException;
import java.util.Random;
import csci3230.hw3.OutputWriter;
public class DoublyLinkedList<Item> implements Iterable<Item> {
private int n; // number of elements on list
private Node pre; // sentinel before first item
private Node post; // sentinel after last item
public DoublyLinkedList() {
pre = new Node();
post = new Node();
pre.next = post;
post.prev = pre;
}
// linked list node helper data type
private class Node {
private Item item;
private Node next;
private Node prev;
}
public boolean isEmpty() {
return (n == 0);
// your code
}
public int size() {
return n;
// your code
}
// add the item to the list
public void add(Item item) {
Node last = post.prev;
Node x = new Node();
x.item = item;
x.next = post;
x.prev = last;
post.prev = x;
last.next = x;
n++;
// your code
}
public ListIterator<Item> iterator() { return new DoublyLinkedListIterator(); }
// assumes no calls to DoublyLinkedList.add() during iteration
private class DoublyLinkedListIterator implements ListIterator<Item> {
private Node current = pre.next; // the node that is returned by next()
private Node lastAccessed = null; // the last node to be returned by prev() or next()
// reset to null upon intervening remove() or add()
private int index = 0;
public boolean hasNext() {
return (index < n); // your code
}
public boolean hasPrevious() {
return (index > 0);
// your code
}
public int previousIndex() {
return (index - 1);
// your code
}
public int nextIndex() {
return (index + 1);
// your code
}
public Item next() {
if (!hasNext()) throw new NoSuchElementException();
lastAccessed = current;
Item item = current.item;
current = current.next;
index++;
return item;
// your code
}
public Item previous() {
if (!hasPrevious()) throw new NoSuchElementException();
current = current.prev;
index--;
lastAccessed = current;
return current.item;
// your code
}
// replace the item of the element that was last accessed by next() or previous()
// condition: no calls to remove() or add() after last call to next() or previous()
public void set(Item item) {
if (lastAccessed == null) throw new IllegalStateException();
lastAccessed.item = item;
// your code
}
// remove the element that was last accessed by next() or previous()
// condition: no calls to remove() or add() after last call to next() or previous()
public void remove() {
if (lastAccessed == null) throw new IllegalStateException();
Node x = lastAccessed.prev;
Node y = lastAccessed.next;
x.next = y;
y .prev = x;
n--;
if (current == lastAccessed) {
current = y;
}
else {
index--;
}
lastAccessed = null;
// your code
}
// add element to list
public void add(Item item) {
Node x = current.prev;
Node y = new Node();
Node z = current;
y.item = item;
x.next = y;
y.next = z;
z.prev = y;
y.prev = x;
n++;
index++;
lastAccessed = null;
// your code
}
}
public String toString() {
StringBuilder s = new StringBuilder();
for (Item item : this)
s.append(item + " ");
return s.toString();
}
}
# 1 楼答案
您可以将弹出的第一个整数保留为临时“最大值”，并将随后弹出的每个整数与之进行比较，以确定是否应替换它。在函数的最末端，弹出所有值并将最大的值添加回函数