我想创建一个称为inBetween的方法，该方法接受一个项目作为参数，如果该项目在最小和最大列表元素之间，则返回true。也就是说，基于为列表元素定义的compareTo方法，该项目大于最小的列表元素，并且小于最大的列表元素。否则，该方法将返回false（即使该项“匹配”最小或最大元素）。
public class DoublyLinkedList {
private Link first; // ref to first item
private Link last; // ref to last item
// -------------------------------------------------------------
public DoublyLinkedList() // constructor
{
first = null; // no items on list yet
last = null;
}
// -------------------------------------------------------------
public boolean isEmpty() // true if no links
{
return first == null;
}
// -------------------------------------------------------------
public void insertFirst(long dd) // insert at front of list
{
Link newLink = new Link(dd); // make new link
if (isEmpty()) // if empty list,
{
last = newLink; // newLink <-- last
} else {
first.previous = newLink; // newLink <-- old first
}
newLink.next = first; // newLink --> old first
first = newLink; // first --> newLink
}
// -------------------------------------------------------------
public void insertLast(long dd) // insert at end of list
{
Link newLink = new Link(dd); // make new link
if (isEmpty()) // if empty list,
{
first = newLink; // first --> newLink
} else {
last.next = newLink; // old last --> newLink
newLink.previous = last; // old last <-- newLink
}
last = newLink; // newLink <-- last
}
// -------------------------------------------------------------
public Link deleteFirst() // delete first link
{ // (assumes non-empty list)
Link temp = first;
if (first.next == null) // if only one item
{
last = null; // null <-- last
} else {
first.next.previous = null; // null <-- old next
}
first = first.next; // first --> old next
return temp;
}
// -------------------------------------------------------------
public Link deleteLast() // delete last link
{ // (assumes non-empty list)
Link temp = last;
if (first.next == null) // if only one item
{
first = null; // first --> null
} else {
last.previous.next = null; // old previous --> null
}
last = last.previous; // old previous <-- last
return temp;
}
// -------------------------------------------------------------
// insert dd just after key
public boolean insertAfter(long key, long dd) {
// (assumes non-empty list)
Link current = first; // start at beginning
while (current.dData != key) // until match is found,
{
current = current.next; // move to next link
if (current == null) {
return false; // didn't find it
}
}
Link newLink = new Link(dd); // make new link
if (current == last) // if last link,
{
newLink.next = null; // newLink --> null
last = newLink; // newLink <-- last
} else // not last link,
{
newLink.next = current.next; // newLink --> old next
// newLink <-- old next
current.next.previous = newLink;
}
newLink.previous = current; // old current <-- newLink
current.next = newLink; // old current --> newLink
return true; // found it, did insertion
}
// -------------------------------------------------------------
public Link deleteKey(long key) // delete item w/ given key
{ // (assumes non-empty list)
Link current = first; // start at beginning
while (current.dData != key) // until match is found,
{
current = current.next; // move to next link
if (current == null) {
return null; // didn't find it
}
}
if (current == first) // found it; first item?
{
first = current.next; // first --> old next
} else // not first
// old previous --> old next
{
current.previous.next = current.next;
}
if (current == last) // last item?
{
last = current.previous; // old previous <-- last
} else // not last
// old previous <-- old next
{
current.next.previous = current.previous;
}
return current; // return value
}
// -------------------------------------------------------------
public void displayForward() {
System.out.print("List (first-->last): ");
Link current = first; // start at beginning
while (current != null) // until end of list,
{
current.displayLink(); // display data
current = current.next; // move to next link
}
System.out.println("");
}
// -------------------------------------------------------------
public void displayBackward() {
System.out.print("List (last-->first): ");
Link current = last; // start at end
while (current != null) // until start of list,
{
current.displayLink(); // display data
current = current.previous; // move to previous link
}
System.out.println("");
}
// -------------------------------------------------------------
public DoublyLinkedList inBetween(long n) {
}
} // end class DoublyLinkedList
////////////////////////////////////
public class InBetweenDemo
{
public static void main(String[] args)
{ // make a new list
DoublyLinkedList theList = new DoublyLinkedList();
theList.insertFirst(22); // insert at front
theList.insertFirst(44);
theList.insertFirst(66);
theList.insertLast(11); // insert at rear
theList.insertLast(33);
theList.insertLast(55);
theList.displayForward();
int n=55;// display list forward
System.out.println("inBetween("+n+") "+ inBetween(n));
theList.displayBackward(); // display list backward
theList.deleteFirst(); // delete first item
n=55;
System.out.println("inBetween("+n+") "+ theList.inBetween(n));
theList.deleteLast();
n=33;
System.out.println("inBetween("+n+") "+ theList.inBetween(n));
theList.deleteKey(22); // delete item with key 11
System.out.println("inBetween("+n+") "+ theList.inBetween(n));
theList.displayForward(); // display list forward
theList.insertAfter(11, 77); // insert 77 after 22
theList.insertAfter(33, 88); // insert 88 after 33
theList.displayForward(); // display list forward
} // end main()
} // end class DoublyLinkedApp
////////////////////////////////////////////////////////////////
我当时想我可以分配一个最大值和一个最小值，然后检查参数是否小于和大于各个值。如果是，那么我将返回true，否则将返回false。我不确定如何开始在无序列表中查找最大值和最小值的代码。
答案 0 :(得分：1)
有两种方法可以解决您的问题。 （可能还有其他方法）
min和max。只需在您的inBetween方法中调用此方法即可。此方法将具有O（n）的最坏情况。 （如果您打算这样做，那么min和max的变量是不够的，则每次调用inBetween时都必须调用该方法，因为可能会值已更新）min和max设置一个变量，然后在每次插入和删除后对其进行更新。它必须是Link类型。在插入中，它将仅具有O（1）的运行时，因为您将直接比较它们的值。在删除时，您将必须比较密钥，如果密钥具有相同的密钥，则必须找到另一个min或max。因此，您还应该创建一种方法来找到min和max。在inBetween方法中，您只需要获取变量min和max。在执行min时，max和inBetween的值将不会更新，因为您要在每次插入和删除时都更新min和max。因此，您可以从两者中选择要实现的目标。
答案 1 :(得分：1)
只需遍历列表，并确保您的商品少于至少一个元素，而又大于另一个：
public boolean inBetween(long n) {
boolean less = false, more = false;
for (Link current = first; current != null; current = current.next)
if ((less |= n < current.dData) & (more |= n > current.dData))
return true;
return false;
}
答案 2 :(得分：0)
我认为您应该首先创建两个变量MAX和MIN。之后，遍历列表并找到MAX和MIN值。因此，选择您想要的值并进行比较。如果该值大于MIN且小于MAX，则为有效数字。我建议您在List的类上添加一个名为listLenght的变量。添加内容时，更新变量listLenght。删除后，请执行相同操作。
答案 3 :(得分：0)
遍历列表以找到min和max的值，如果输入值大于min且小于max，则返回true：
public static boolean isBetween(List<Integer> list, int value){
int min = Integer.MAX_VALUE, max = Integer.MIN_VALUE;
for (int i : list){
if (i < min)
min = i;
if (i > max)
max = i;
}
return value > min && value < max;
}