[英]How to find Intersections in two singly linked lists
我正在编写一个程序,其中我必须设置数据结构字典(单链表),按字母顺序排列单词(出现在带有文档ID的文本文档中的句子中的单词)。 并找出哪些单词出现在多个文档中,以便教授希望我们做一个交集。 我真的很困惑如何做交叉。 我有其他一切(我相信这是正确的)。 这是我的代码(我添加了我的交叉算法,但它显然不起作用,我遵循教授算法[她从未向我们展示过一个例子]):
public class dictionary
{
//variables
dNode head;
int size;
//constructor
public dictionary()
{
head = null;
size = 0;
}
//addFirst method
public void addFirst(dNode s)
{
s.setNext(head);
head = s;
size++;
}
public void addLast(dNode s)
{
if ( head == null )
{
head = s;
}
else
{
s.setNext(null);
dNode w = head;
while ( w.getNext() != null )
{
w = w.getNext();
}
w.setNext(s);
}
size++;
}
//toString Method
public String toString()
{
String w = "";
dNode s = head;
while ( s != null )
{
w += s + "\n";
s = s.getNext();
}
return w;
}
//intersection method
public String intersection(pNode head, dNode head) {
int left = posting.head;
int right = dictionary.head;
int result = new dictionary();
while (left != null && right != null) {
if (dID.left < dID.right) {
left = left.next;
else if (dID.left > dID.right)
right = right.next;
else
left = left.next;
right = right.next;
result.push(left.data() );
}
}
return result;
}
}
public class dNode
{
//variables
String sent;
posting post;
dNode nextNode;
//constructor
public dNode(String sent, posting post, dNode nextNode)
{
this.sent = sent;
this.post = post;
this.nextNode = nextNode;
}
//returns element of this node
public String getSent() {
return sent;
}
//retunrs the next node of this node
public dNode getNext() {
return nextNode;
}
//modifier methods
//sets elements of this node.
public void setSent(String newSent) {
sent = newSent;
}
//sets the next node of this node
public void setNext( dNode newNext) {
nextNode = newNext;
}
//toString method
public String toString()
{
return "Sentence and Posting: \n" + sent + "\n" + post;
}
}
public class pNode {
//variables
int dID;
String word;
int occurence;
pNode next;
//constructor
public pNode(int dID, String word, int occurence, pNode next)
{
this.dID = dID;
this.word = word;
this.occurence = occurence;
this.next = next;
}
//return element of this node
public String getWord() {
return word;
}
//Returns the next node of this node
public pNode getNext() {
return next;
}
//Modifier methods
//set the words of this node
public void setWord(String newWord) {
word = newWord;
}
//sets the next node of this node
public void setNext(pNode newNext){
next = newNext;
}
//toString method
public String toString() {
return "Document ID, Word, Occurence: \n " + dID + ", "
+ word + ", " + occurence;
}
}
public class posting
{
//variables
pNode head;
int size;
//constructor
public posting()
{
head = null;
size = 0;
}
//addFirst method
public void addFirst(pNode s)
{
s.setNext(head);
head = s;
size++;
}
//addLast method
public void addLast(pNode s)
{
if ( head == null )
{
head = s;
}
else
{
s.setNext(null);
pNode w = head;
while ( w.getNext() != null )
{
w = w.getNext();
}
w.setNext(s);
}
size++;
}
//toString method
public String toString()
{
String w = "";
pNode s = head;
while ( s != null)
{
w += s + "\n";
s = s.getNext();
}
return w;
}
}
import java.io.*;
import java.util.*;
public class testFile
{
public static void main (String[] args) throws FileNotFoundException
{
File filename = new File("/export/home/hawkdom2/s0878044/CS503/assignment2/sentences.txt");
Scanner scan = new Scanner(filename);
dictionary Dictionary = new dictionary();
while ( scan.hasNextLine() )
{
String sentence = scan.nextLine();
String[] word = sentence.split(" ");
//first element is document id
int dID = Integer.parseInt( word[0] );
//insertion sort
for ( int i = 2; i < word.length; i++ )
{
for ( int j = i; j > 1; j-- )
{
if ( word[j].compareTo( word[j-1] ) > 0 )
{
String switchs = word[j];
word[j] = word[j-1];
word[j-1] = switchs;
}
}
}
//integer array count
int[] count = new int[word.length];
for ( int i = 1; i < word.length; i++)
{
for ( int j = 1; j < word.length; j++)
{
if (word[i].equalsIgnoreCase( word[j] ) )
{
count[i]++;
}
}
}
posting posts = new posting();
for ( int i = 1; i < word.length; i++ )
{
if ( (i > 1 ) && (word[i].equalsIgnoreCase( word[i-1] ) ) )
continue;
else
{
posts.addFirst(new pNode(dID, word[i], count[i], null) );
}
}
Dictionary.addLast(new dNode(sentence, posts, null) );
}
//print out output
System.out.println(Dictionary);
}
}
这是句子文件:
1 a rose is a rose
2 John chased a cat and the cat chased John
3 cats are mammals but mammals are not cats
4 beavers build dams but i know a beaver that does not
5 my dog chased a cat and the cat attacked my dog
6 my dog likes cats but my cat dislikes dogs
7 my dog likes roses but roses dislike my dog
8 my cat dislikes roses but roses like my cat
9 red roses are not my favorite roses
10 my favorite roses are pink roses
如果我能够了解如何交叉两个链接列表(或者如果我的程序有任何其他错误),我会非常感激。 我上周病了,我的教授拒绝帮助我做错过的事情(显然,如果我生病时不上课,我不是一个认真的程序员)。 我真的不能忍受这位教授教授这门课程的方式,因为她没有给我们任何程序的例子(而且她给我们的极少数总是有错误)。 她也只是给我们算法,她已经说过,但它们并不总是正确的。 我曾经喜欢编程,但她真的让我失望了,我现在要做的就是至少得到一个C所以我可以切换到IT。 如果有人可以帮助我,我真的很感激,我非常渴望完成这门课程而不必再接受这位教授。
我添加了一个交叉方法,但仍然收到所有这些错误:找到7个错误:文件:/export/home/hawkdom2/s0878044/CS503/assignment2/testFile.java [line:86]错误:/ export / home / hawkdom2 / s0878044 /CS503/assignment2/testFile.java:86:非法启动表达式文件:/export/home/hawkdom2/s0878044/CS503/assignment2/testFile.java [line:86]错误:/ export / home / hawkdom2 / s0878044 / CS503 /assignment2/testFile.java:86:';' 预期文件:/export/home/hawkdom2/s0878044/CS503/assignment2/testFile.java [line:86]错误:/export/home/hawkdom2/s0878044/CS503/assignment2/testFile.java:86:不是声明文件: /export/home/hawkdom2/s0878044/CS503/assignment2/testFile.java [line:86]错误:/export/home/hawkdom2/s0878044/CS503/assignment2/testFile.java:86:';' 预期文件:/export/home/hawkdom2/s0878044/CS503/assignment2/testFile.java [line:86]错误:/export/home/hawkdom2/s0878044/CS503/assignment2/testFile.java:86:不是声明文件: /export/home/hawkdom2/s0878044/CS503/assignment2/testFile.java [line:86]错误:/export/home/hawkdom2/s0878044/CS503/assignment2/testFile.java:86:';' 预期文件:/export/home/hawkdom2/s0878044/CS503/assignment2/testFile.java [line:96]错误:/export/home/hawkdom2/s0878044/CS503/assignment2/testFile.java:96:'其他'没有'如果'
public static void findIntersection(LLNode head1, LLnode head2)
{
HashSet<LLNode> hs= new HashSet<>();
HashSet<LLNode> hs2 = new HashSet<>();
LLNode currentNode1 = head1;
while(currentNode.getNext()!=null)
{
hs.add(currentNode);
currentNode1 = currentNode1.getNext();
}
LLNode currentNode2 = head2;
while(currentNode2.getNext()!=null)
{
if(hs1.contains(currentNode2)){
hs2.add(currentNode2);
}
}
相交两个排序列表很容易。
从指向每个列表中第一个节点的指针开始。 让我们称他们为left
和right
。 创建一个新的空列表,让我们称之为result
。 现在你循环,将存储在数据left
和right
节点:
left
的数据较少,则left
前进。 right
的数据较少,则right
前进。 result
和推进两left
和right
。 你的循环继续,直到你读到left
或right
的结尾。 现在您已经交叉了两个列表( 即 result
列表仅包含left
right
出现的元素)。
function intersect( list1, list2 )
left <- list1.head
right <- list2.head
result <- new list
while left != null and right != null
if left.data < right.data
left <- left.next
elseif left.data > right.data
right <- right.next
else
left <- left.next
right <- right.next
result.push(left.data)
end if
end
return result
end function
为了获得最佳速度,可以在列表中存储tail
指针(以便addLast
快速),或者始终在前面添加(因此您的结果是反向排序的)然后反转列表(使用线性时间的简单操作 -复杂)。
我的版本找到两个链表的交集如下:1。迭代第一个链表并在IdentityHashMap中添加元素作为键。 2.迭代第二个链表并进行containsKey检查。 如果为true,则节点相交。
时间复杂度:O(n + m)其中n =第一个列表的大小,m =第二个列表的大小空间复杂度:O(n)
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