双重链接列表没有显示我正在搜索的确切值

Question

我正在使用searchByName方法在Doubly LinkedList中搜索节点的精确值即使我传递了LinkedList中存在的值，它也不会显示我想要的数据。

public void searchByName(String param) throws Exception{
Node currentNode = start;
String theFirstName = currentNode.firstName.toLowerCase();
String theLastName = currentNode.lastName.toLowerCase();
param = param.toLowerCase();
if (start == null) {
throw new Exception("List Underflow");
}else{
String id= "Student ID", ln="Last Name", fn="First Name", course="Course", section="Section", yl="Year Level";
System.out.format("%-10st%-10st%-10st%-5st%-10st%s", id, ln, fn, course, section, yl);
System.out.println();
while(currentNode.next != null){
if (param == theFirstName || param == theLastName) {
System.out.format("%-10st%-10st%-10st%-5st%-15st%d", currentNode.studentID, currentNode.lastName, currentNode.firstName, currentNode.course, currentNode.section, currentNode.yearLevel);
System.out.println();
}else{
System.out.println("Not found");
break;
}
currentNode = currentNode.next;
}
if (currentNode.next == null){
System.out.format("%-10st%-10st%-10st%-5st%-15st%d", currentNode.studentID, currentNode.lastName, currentNode.firstName, currentNode.course, currentNode.section, currentNode.yearLevel);
System.out.println();
}
}

我的主要功能：

public static void main(String[] args)  throws Exception {
StudentRecord senators = new StudentRecord();
senators.insertEnd("110007", "Lacson", "Ping", "BSCS", "BSCS-III-A", "Active",  3);
senators.insertEnd("110008", "Angara", "Sonny", "BSCS", "BSCS-III-B", "InActive",  3);
senators.searchByName("Lacson");
}

要点链接：https://gist.github.com/30b27d3612f95fc2ced99f50c4f23c14

Answer 1

您的方法中有许多错误2个主要错误：

1. 字符串应该与equals方法进行比较，而不是==
2. 您遍历列表的算法错误
3. 始终使用您自己的异常(LinkedListOutOfBoundsException(
4. 不要修改函数内部的输入参数
5. 由于抛出了不必要的else语句
6. 最后一个if是绝对无用的
7. 尝试使用记录器

public void searchByName(String param) throws LinkedListOutOfBoundsException {
if (null == start) {
throw new LinkedListOutOfBoundsException("List Underflow");
}
if (null == param) {
throw new IllegalArgumentException("param must not be null");
}
Node currentNode = start;    
while (currentNode != null) {
if (param.equalsIgnoreCase(currentNode.firstName) 
|| param.equalsIgnoreCase(currentNode.lastName)) {
break;
}
currentNode = currentNode.next;
}
if (null == currentNode) {
LOGGER.info("Not found");
} else {
LOGGER.info("Found {}", param);
}
}

Answer 2

更改节点时，不会更新FirstName和LastName。

另外，将字符串与equals进行比较，而不是==。