java中2个循环单链表并集的方法
[英]A method to get the union of 2 circular singly linked lists in java
问题描述
我在考试前练习了一些链表方法，我试图得到 2 个循环链表的并集，但是在 output 中，元素应该被排序。 我写了代码，但我得到了错误的答案。
我在一个双向链表中尝试了这种方法，它工作得非常好，但当然我必须更改一些符号。
输入：列表：8 | 2 | 4 | 3 | 10 | 5 | 8 | 4 |
列表 1：5 | 1 | 3 | 7 | 0 | 6 | -4 | 1 |
output：5 | 1 | 3 | 7 | 0 | 6 | -4 | 8 | 2 | 4 | 3 | 10 | 5 | 8 | 1 | 5 | 1 | 3 | 7 | 0 | 6 | -4 |
正如你所看到的，我得到的数字比我已经拥有的多。
所以任何帮助将不胜感激。
这是方法：
public CircularLinkedList union(CircularLinkedList a, CircularLinkedList b) {
Element curA = a.head;
Element curB = b.head;
CircularLinkedList c = new CircularLinkedList();
do {
if(curA.data < curB.data) {
c.insert(curA.data);
curA = curA.next;
}else {
c.insert(curB.data);
curB = curB.next;
}
}while(curA != a.rear && curB != b.rear);
do {
c.insert(curA.data);
curA = curA.next;
}while(curA != a.rear);
do {
c.insert(curB.data);
curB = curB.next;
}while(curB != b.rear);
return c;
}
这是整个 class：
public class CircularLinkedList {
class Element{
int data; // int type used as example
Element next; // reference of the successor
Element(int value) {
this.data = value;
this.next = this;
}
}
private Element head = null;
private Element rear = null;
public CircularLinkedList() {
this.head = this.rear = null;
}
public boolean isEmpty() {
return head == null;
}
public boolean findValue(int value) {
Element cur = this.head;
while(cur != null) {
if (cur.data == value)
return true;
cur = cur.next;
}
return false;
}
public int countValue(int value) {
int c = 0; // counter
Element cur = this.head;
if(cur == null)
return 0;
do {
if(cur.data == value)
c++;
cur = cur.next;
}while (cur != this.head);
return c;
}
public boolean isInList(int value) {
if(this.head == null)
return false;
Element cur = this.head;
while(cur != this.rear) {
if(cur.data == value)
return true;
cur = cur.next;
}
return false;
}
@Override
public String toString() {
String str = "";
Element cur = this.head;
if(cur == null)
return "The list is empty";
do {
str += cur.data + " | ";
cur = cur.next;
}while (cur != this.head);
return str;
}
public void insert(int value) {
Element tmp = new Element (value);
//special case: empty list
if(this.head == null) {
tmp.next = tmp;
this.head = tmp;
this.rear = tmp;
}else { // general case
tmp.next = this.head;
this.rear.next = tmp;
this.rear = tmp;
}
}
public void deleteAtHead() {
if(this.head == null)
return;
Element cur = this.head;
while(cur.next != this.head) {
cur = cur.next;
}
cur.next = cur.next.next;
this.head = this.head.next;
return ;
}
public void deleteAtRear() {
if(this.head == null)
return;
Element cur = this.head;
// Element prev = null;
while(cur.next != this.rear) {
// prev = cur;
cur = cur.next;
}
cur.next = cur.next.next;
this.rear = cur;
}
public boolean delete(int value) {
Element cur = this.head;
if(this.head.data == value) { //if the node to be deleted is head node
while(cur.next != this.head) { //iterate till the last node i.e. the node which is pointing to head
cur = cur.next;
}
cur.next = cur.next.next; // update current node pointer to next node of head
this.head = this.head.next; //update head node
return true;
}
else { // if node to be deleted is other than head node
Element prev = cur; // track previous node from current (node)
while(cur.data != value) { // find the node
prev = cur;
cur = cur.next;
}
prev.next = cur.next; //updating next field of previous node to next of current node.current node deleted
return true;
}
}
public void deleteEven() {
// if(this.head == null)
// return;
//
// //case of deleting the head
// if(this.head.data % 2 == 0) {
// this.head.next = this.head;
// this.rear.next = this.head;
// if(this.head == null)
// this.rear = null;
// }
//
// Element cur = this.head;
// Element prev = cur;
// while(cur != this.head) {
// prev = cur;
// cur = cur.next;
// }
// prev.next = cur.next;
if(this.head == null)
return;
Element cur = this.head;
while(cur.next != this.head) {
if(cur.data % 2 == 0)
this.delete(cur.data);
cur = cur.next;
}
}
public void deleteLastOccurence(int value) {
Element cur = this.head;
Element prev = null;
Element tmp = null;
if(this.head == null)
return;
if(this.rear.data == value) {
this.head = null;
return;
}
// while(cur != this.rear) {
while(cur.next != this.head && cur.next.data != value) {
prev = cur;
tmp = cur.next;
// cur = cur.next;
}
// cur = cur.next;
// }
prev.next = tmp.next;
}
public void deleteAllOccurrences(int value) {
Element cur = this.head;
Element next = null;
if (cur.data == value) {
cur = cur.next;
this.head = cur;
}
do {
next = cur.next;
if (next.data == value) {
cur.next = next.next;
}
cur = next;
} while (cur != this.head);
}
public CircularLinkedList union(CircularLinkedList a, CircularLinkedList b) {
Element curA = a.head;
Element curB = b.head;
CircularLinkedList c = new CircularLinkedList();
do {
if(curA.data < curB.data) {
c.insert(curA.data);
curA = curA.next;
}else {
c.insert(curB.data);
curB = curB.next;
}
}while(curA != a.rear && curB != b.rear);
do {
c.insert(curA.data);
curA = curA.next;
}while(curA != a.rear);
do {
c.insert(curB.data);
curB = curB.next;
}while(curB != b.rear);
return c;
}
public CircularLinkedList inter(CircularLinkedList a, CircularLinkedList b) {
Element curA = a.head;
CircularLinkedList c = new CircularLinkedList();
if(a.head == null || b.head == null)
return c;
while(curA != a.rear) {
if(b.isInList(curA.data))
c.insert(curA.data);
curA = curA.next;
}
return c;
}
public boolean isEqualTo(CircularLinkedList a) { //same content in same order
Element cur = this.head;
Element curA = a.head;
while(cur != this.rear && curA != a.rear) {
if(cur.data != curA.data) {
return false;
}
cur = cur.next;
curA = curA.next;
}
return true;
}
public int countOddNbrs() {
if(this.head == null)
return 0;
int c = 0;
Element cur = this.head;
do {
if(cur.data % 2 != 0)
c++;
cur = cur.next;
}while(cur != this.head);
return c;
}
// public int findLastOccurence(int value) {
//
// }
public static void main(String[] args) {
CircularLinkedList list = new CircularLinkedList();
CircularLinkedList list1 = new CircularLinkedList();
CircularLinkedList list2 = new CircularLinkedList();
list.insert(8);
list.insert(2);
list.insert(4);
list.insert(3);
list.insert(10);
list.insert(5);
list.insert(8);
list.insert(4);
System.out.println(list);
// list2.insert(8);
// list2.insert(2);
// list2.insert(4);
// list2.insert(3);
// list2.insert(10);
// list2.insert(5);
// list2.insert(8);
// list2.insert(4);
// System.out.println(list);
list1.insert(5);
list1.insert(1);
list1.insert(3);
list1.insert(7);
list1.insert(0);
list1.insert(6);
list1.insert(-4);
list1.insert(1);
System.out.println(list1);
// System.out.println(list.findValue(2)); // working
// list.delete(8); // working
// System.out.println(list);
// System.out.println(list.countOddNbrs()); //working
// list.deleteEven(); // working
// System.out.println(list);
// list.deleteAtHead(); // working
// System.out.println(list);
// list.deleteAtRear(); // working
// System.out.println(list);
// list.deleteLastOccurence(4); //not working
// System.out.println(list);
// list.deleteAllOccurrences(4); // working
// System.out.println(list);
System.out.println(list2.union(list, list1)); //not working
// System.out.println(list2.inter(list, list1)); //working
// System.out.println(list.isEqualTo(list1)); // working
}
}
1 个解决方案
解决方案1
1 已采纳 2021-05-15 19:19:57
这应该这样做。 虽然这只会在两个输入都排序的情况下产生一个排序列表。
public CircularLinkedList union(CircularLinkedList a, CircularLinkedList b) {
Element curA = a.head;
Element curB = b.head;
CircularLinkedList c = new CircularLinkedList();
do {
if (curA != null && (curB == null || curA.data < curB.data)) {
c.insert(curA.data);
curA = curA.next === a.rear ? null : curA.next;
} else {
c.insert(curB.data);
curB = curB.next === b.rear ? null : curB.next;
}
} while(curA != null || curB != null);
return c;
}
要保持列表排序，请调整您的插入方法：
public void insert(int value) {
Element tmp = new Element(value);
Element cur = head;
// special case: empty list
if (this.head == null) {
tmp.next = tmp;
this.head = tmp;
this.rear = tmp;
} else { // general case
// Change this condition if you want the list to be sorted decending
while (cur != rear && cur.data < value) {
prev = cur;
cur = cur.next;
}
tmp.next = cur;
cur.next = tmp;
}
}
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