本文介紹了刪除循環(huán)鏈表中的所有匹配項(xiàng)。(Java)的處理方法，對(duì)大家解決問(wèn)題具有一定的參考價(jià)值，需要的朋友們下面隨著小編來(lái)一起學(xué)習(xí)吧！
問(wèn)題描述
我要實(shí)現(xiàn)一個(gè)方法，該方法刪除作為參數(shù)給定的特定值的所有匹配項(xiàng)。
我創(chuàng)建了2個(gè)臨時(shí)元素，并將它們引用到頭部，我將使用它們?yōu)g覽列表。我能夠編寫(xiě)完整的代碼，但沒(méi)有得到任何結(jié)果，我不知道問(wèn)題出在哪里，也沒(méi)有收到錯(cuò)誤。
如有任何幫助，我們將不勝感激。
這是方法(我的方法)：
public void deleteAllOccurrences(int value) {
Element cur = this.head;
Element prev = null;
//deleting head
while(cur != this.rear && cur.data == value) {
cur.next = this.head;
cur = this.head;
}
while(cur != this.rear) {
while(cur != this.rear && cur.data != value) {
prev = cur;
cur = cur.next;
}
if(cur == this.rear)
return;
prev.next = cur.next;
cur = prev.next;
}
}
這是整個(gè)類，您可能需要檢查一些內(nèi)容：
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;
}
@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) {
this.head = tmp;
}else { // general case
tmp.next = this.head.next;
this.head.next = tmp;
}
}
public void deleteAtHead() {
if(this.head == null) {
return;
}
else {
if(this.head != this.rear) {
this.head = this.head.next;
this.rear.next = this.head;
}
else {
this.head = this.rear = null;
}
}
}
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 != this.rear) {
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.head.data == value) {
this.head = null;
return;
}
while(cur != this.rear) {
if(cur.next != null && cur.next.data == value) {
prev = cur;
tmp = cur.next;
}
cur = cur.next;
}
prev.next = tmp.next;
}
public void deleteAllOccurrences(int value) {
Element cur = this.head;
Element prev = null;
//deleting head
while(cur != this.rear && cur.data == value) {
cur.next = this.head;
cur = this.head;
}
while(cur != this.rear) {
while(cur != this.rear && cur.data != value) {
prev = cur;
cur = cur.next;
}
if(cur == this.rear)
return;
prev.next = cur.next;
cur = prev.next;
}
}
// public CircularLinkedList union(CircularLinkedList a, CircularLinkedList b) {
//
// }
//
//
// public CircularLinkedList inter(CircularLinkedList a, CircularLinkedList b) {
//
// }
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();
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);
// System.out.println(list.findValue(2)); // working
// list.delete(2); // working
// System.out.println(list);
// System.out.println(list.countOddNbrs()); //working
// list.deleteEven(); // not working
// System.out.println(list);
// list.deleteAtHead(); // not working
// System.out.println(list);
// list.deleteLastOccurence(4); //not working
// System.out.println(list);
list.deleteAllOccurrences(4);
System.out.println(list);
}
}
推薦答案
您的方法正在無(wú)限循環(huán)中運(yùn)行。試著這樣做：
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);
}
當(dāng)頭部是可拆卸的元素時(shí)，您也應(yīng)該處理邊角情況。
這篇關(guān)于刪除循環(huán)鏈表中的所有匹配項(xiàng)。(Java)的文章就介紹到這了，希望我們推薦的答案對(duì)大家有所幫助，
