正在研究我的双向链表的搜索方法。我遇到了异常，但我似乎无法弄清楚如何在不获取异常的情况下遍历列表..
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();
}
}
}
最佳答案
我没有你的行数，所以我在这里盲目猜测，但我会猜测这一行是你的问题:
if (head.getFirstName().contains(searchEntry) || head.getLastName().contains(searchEntry) || head.getPhoneNum().contains(searchEntry) || head.getEmail().contains(searchEntry))
您的其中一个条目很可能返回 null对于 getFirstName , getLastName , getPhoneNum ，或getEmail .
您必须检查每个不是 null在解除引用之前(你不能这样做 null.someMethod() )
执行此操作的一种方法:
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)))
关于java - 双向链表查找法，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/9559904/