用单链表保存m个整数，结点的结构为（data，next）且|data|<=n（n为正整数）。现在要求设计一个时间复杂度尽可能高效的算法，对于链表中data的绝对值相等的结点，仅保留第一次出现的结点而删除其余绝对值相等的结点。
public class E2 {
static class Node {
int data;
Node next;
}
private static Node createLinked() {
System.out.println("请输入结点个数");
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
Node head = new Node();
head.next = null;
Node r = head;
while (n-- > 0) {
Node e = new Node();
e.data = scanner.nextInt();
e.next = null;
r.next = e;
r = e;
}
return head;
}
private static Node deleteRepeatNode(Node l) {
//获得首元结点
Node p = l.next;
while (p != null) {
//用于获得待删除结点的前驱结点
Node p1 = p;
Node p2 = p.next;
while (p2 !=null) {
if (p2.data == p.data) {
p1.next = p2.next;
p2 = p2.next;
}
else {
p2 = p2.next;
p1 = p1.next;
}
}
p = p.next;
}
return l;
}
public static void main(String[] args) {
Node linked = createLinked();
Node node = deleteRepeatNode(linked);
Node p = node.next;
while (p != null) {
System.out.println(p.data);
p = p.next;
}
}
}
