我已经创建了一个已排序的链接列表,现在我正在尝试找出如何删除重复项。我想添加在我创建的Add方法中执行此操作的代码,但我似乎无法弄明白。我觉得这应该相对容易,但我现在有点脑死了。
在我的添加方法中,我检查索引以查看要添加项目的位置。 “索引”是一个int变量,但我想检查“item”,一个可比较的,是否是它之前存储的相同项目。我想使用compareTo方法,但我会得到一个类型不匹配。有没有人知道更好的方法呢?
以下是我的添加方法的代码:
package sortedListReferenceBased;
public class SortedListReferenceBasedIterativeNoDuplicates
implements SortedListInterface {
// reference to linked list of items
private Node head;
private int numItems; // number of items in list
public SortedListReferenceBasedIterativeNoDuplicates() {
numItems = 0;
head = null;
} // end default constructor
public boolean sortedIsEmpty() {
return numItems == 0;
//TODO
} // end sortedIsEmpty
public int sortedSize() {
return numItems;
//TODO
} // end sortedSize
private Node find(int index) {
// --------------------------------------------------
// Locates a specified node in a linked list.
// Precondition: index is the number of the desired
// node. Assumes that 1 <= index <= numItems+1
// Postcondition: Returns a reference to the desired
// node.
// --------------------------------------------------
Node curr = head;
for (int skip = 1; skip < index; skip++) {
curr = curr.getNext();
} // end for
return curr;
} // end find
public Comparable sortedGet(int index)
throws ListIndexOutOfBoundsException {
if (index >= 1 && index <= numItems){
Node curr = find(index);
Object dataItem = curr.getItem();
return (Comparable) dataItem;
}
else {
throw new ListIndexOutOfBoundsException("List index out of bounds on get.");
}
//TODO
} // end sortedGet()
public void sortedAdd(Comparable item) throws ListException{
int index = locateIndex(item); //to find location where item should be added
if( index >=1 && index <= numItems+1){
//if adding an item to the very beginning of list
if (index == 1){
Node newNode = new Node(item,head);
head = newNode;
}
if (item.compareTo(something something?)== 0){ //if item is a duplicate of previous item do nothing
System.out.println("No duplicates!");
}
//advances
else {
Node prev = find(index-1); //finds out where previous node is
Node newNode = new Node(item, prev.getNext()); //creates Node with item you wish to add
prev.setNext(newNode); //links new node with previous node
}
numItems++;
}//end main if statement
else {
throw new ListIndexOutOfBoundsException("List index out of bounds on add.");
}
//TODO
} // end sortedAdd()
public void sortedRemove(Comparable item) throws ListException {
int index = locateIndex(item);
if (index >= 1 && index <= numItems){ //if the index is greater than 1 (meaning list not empty) and
//index doesn't exceed list size do the following:
//if index is value of one then delete first node in this special way
if (index == 1) {
head = head.getNext();
}
//if there is only one item in the list then set head to nothing so index out of bounds error won't occur
if (numItems == 1){
head = null;
}
else { //if none of these things occur go ahead and delete item, allocating Nodes accordingly
Node prev = find(index-1);
Node curr = prev.getNext();
prev.setNext(curr.getNext());
}
numItems--;//must account for one less item
}
if (!sortedIsEmpty()){
System.out.println("Item does not exist!");
}
else { //if index doesn't meet if statement requirements
throw new ListIndexOutOfBoundsException("List index out of bounds on remove.");
}
//TODO
} // end sortedRemove
public void sortedRemoveAll() {
// setting head to null causes list to be
// unreachable and thus marked for garbage
// collection
head = null;
numItems = 0;
} // end sortedRemoveAll
//Returns the position where item belongs or exists in a sorted list;
//item and the list are unchanged.
public int locateIndex(Comparable item) {
Node curr = head;
for (int i = 1; i <= sortedSize(); i++){
if (item.compareTo(curr.getItem())<= 0){
return i;
}//end if
else {
curr = curr.getNext();
}//end else
}//end for
return sortedSize()+1;
//TODO
} //end locateIndex()
} // end ListReferenceBased
我为奇怪的格式道歉。现在非常粗糙。如果这个问题非常明显,我也很抱歉!哈哈
答案 0 :(得分:4)
初步要点:
我不明白为什么你似乎试图在Java中实现链表...因为已经有java.util.LinkedList
形式的完美的实现。
没有重复的集合是一组......
基于链表的集合将不是最理想的。例如,对于基于树的实现,插入O(N)
与O(logN)
进行比较,对于基于散列表的实现,插入O(1)
(假设其大小适当)。 java.util.TreeSet
和java.util.HashSet
分别是示例。
话虽如此,假设你真的需要洞察力/暗示......
如果您有预先排序的链接列表,那么删除重复项的方法是逐步执行节点,将node.value
与node.next.value
进行比较。如果值相等,则表示您找到了重复项,您可以将node.next
更改为node.next.next
来删除它。您的代码还需要处理各种“边缘情况”;例如列表包含0或1个元素等
答案 1 :(得分:0)
您是否已开始使用链接列表?使用内置的TreeSet似乎更适合这种情况。
答案 2 :(得分:0)
尝试
if (locateIndex(item) != (sortedSize() + 1)) { //locateIndex returns sortedSize() + 1 if it didn't find the item, so we check that
System.out.println("No duplicates!");
}
所有关于使用您已编写的代码。