我使用此刪除方法刪除列表的第3個元素。但它會刪除指定索引之前的所有元素。所以如果Jhon,Sam,Philip,Emma是列表元素,如果我刪除第三元素,那麼剩下的元素就是Nick。我怎樣才能解決這個問題?刪除索引函數刪除索引前的所有元素
import java.util.*;
class List {
Customer listPtr;
int index;
public void add(Customer customer) {
Customer temp = customer;
if (listPtr == null) {
listPtr = temp;
index++;
} else {
Customer x = listPtr;
while (x.next != null) {
x = x.next;
}
x.next = temp;
index++;
}
}
public void remove(int index) {
int size = size();
Customer tmp = listPtr, tmp2;
int i = 0;
while (i != size) {
if ((i + 1) == index) {
tmp2 = tmp;
listPtr = tmp2.next;
break;
}
tmp = tmp.next;
++i;
}
}
public int size() {
int size = 0;
Customer temp = listPtr;
while (temp != null) {
temp = temp.next;
size++;
}
return size;
}
public void printList() {
Customer temp = listPtr;
while (temp != null) {
System.out.println(temp);
temp = temp.next;
}
}
}
class DemoList {
public static void main(String args[]) {
List list = new List();
Customer c1 = new Customer("10011", "Jhon");
Customer c2 = new Customer("10012", "Sam");
Customer c3 = new Customer("10013", "Philip");
Customer c4 = new Customer("10014", "Emma");
list.add(c1);
list.add(c2);
list.add(c3);
list.add(c4);
list.remove(3);
System.out.println(list.size());
list.printList();
}
}
class Customer {
String id;
String name;
Customer next;
public Customer(String id, String name) {
this.id = id;
this.name = name;
}
public String toString() {
return id + " : " + name;
}
public boolean equals(Object ob) {
Customer c = (Customer) ob;
return this.id.equals(c.id);
}
}
誰是尼克?... – NPE
你應該考慮保持跟蹤的大小,而不是每次計算它。這是一個O(n)操作,因此您的列表中的每個方法現在都是O(n)。您在add()上增加的索引應該是大小。 –