Question

使用我的双向链表的搜索方法。我得到了例外，但我似乎无法弄清楚如何在没有得到它们的情况下遍历列表..

    public void searchEntryNode() {
    System.out.println("I'll search through each entry to pull up what you're looking for ");
    System.out.println("Type in what you want ");
    String searchEntry = keyboard.next();

    EntryNode n = head;
    while (head != null) {
        if (head.getFirstName().contains(searchEntry) || head.getLastName().contains(searchEntry) || head.getPhoneNum().contains(searchEntry) || head.getEmail().contains(searchEntry)) { 
            System.out.println("Found a matching entry");
            System.out.println(n.getFirstName() + " " + n.getLastName() + " " + n.getEmail() + " " + n.getPhoneNum());
        }

        if (head.getNext() != null) {
            head = head.getNext();
        }   
        else {
                System.out.println("That's all we found ");
                System.out.println();
                menu();
        }
    }
}

Answer 1

我没有你的行数，所以我在这里盲目猜测，但我猜这个行是你的问题：

if (head.getFirstName().contains(searchEntry) || head.getLastName().contains(searchEntry) || head.getPhoneNum().contains(searchEntry) || head.getEmail().contains(searchEntry))

您的某个条目很可能会为null，getFirstName，getLastName或getPhoneNum返回getEmail。

在解除引用之前，你必须检查每个null（你不能null.someMethod()）

这样做的一种方法：

string firstName = head.getFirstName();
string lastName = head.getLastName();
string phoneNum = head.getPhoneNum();
string email = head.getEmail();

if ((firstName != null && firstName.contains(searchEntry)) 
     || (lastName != null && lastName.contains(searchEntry)) 
     || (phoneNum != null && phoneNum.contains(searchEntry)) 
     || (email != null && email.contains(searchEntry)))

双链表搜索方法

1 个答案: