Question

我有一个 * ，我正在尝试使用remove方法删除并返回特定的目标项。我尝试了许多不同的尝试使它工作，但它一直给我NPE。

这是我的第一个删除（）：

这是我的第二个remove（）能够使代码编译：

这是我的LinearNode：

学生班：

Answer 1

删除应该相当简单，你已经有了一般的想法：

public Student remove(Student items) {
    LinearNode  previous = null,
                current = head;

    // iterate over all the nodes starting at the head, maintaining a reference to the previous node as you go          
    while (current != null && current.items.compareTo(items) != 0) {
        previous = current;
        current = current.next;
    }

    // At this point you have either a) found the Node with matching items or b) not found it
    if (current == null) {
        // not found in the list
        return null;
    }

    // At this point you know where the Node is, and you have a reference previous node as well 
    // so it's easy to reattach the linked list to remove the node
    if (previous == null) {
        // The head node was the match if previous is not set, so make sure to update the head Node accordingly
        head = current.next;
    }
    else {
        previous.next = current.next
    }

    return current.items;
}

Answer 2

对可疑代码使用try catch，直到找到导致问题的行。

这会告诉您是否正在寻找合适的地方。

例如：

  while (current != null) {
     try{//1st level try
        if(current.items.compareTo(items) == 0) {
          try{ //2nd level try 
            if(previous == null) {
                head = head.next;
                return items;
            }
            else {
                previous.next = current.next;
                return items;
            }
          }catch(NullPointerException e ){
             StackTraceElement t = e.getStackTrace()[0];  
             System.out.println("catch lvl 2 at line: " + t.getLineNumber());
          }
        }
        else {
            previous = current;
            current = current.next;
        }
      }catch(NullPointerException e){
         StackTraceElement t = e.getStackTrace()[0]; 
         System.out.println("catch lvl 1 at line: " + t.getLineNumber());
      }
    }

修改

您可以尝试将此“try catch”并包装所有主要功能：

try{ ... }catch(NullPointerException e ){ int i=0 for( StackTraceElement t : e.getStackTrace()){ System.out.println("stack[" + i + "]: " + t.getLineNumber()); i++; } }

<强> EDIT2

public Student remove(Integer studentId) { LinearNode previous = null; LinearNode current = head; while (current != null) { //if everything is OK you can remove the 2 ifs if(current.items == null){ //something is really wrong on insert } else if(current.items.getId() == null){ //something is really wrong on insert } else if( studentId.compareTo(curent.items.getId()) == 0) { //return value is curent.items; if(previous == null) { head = head.next; return curent.items; } else { previous.next = current.next; return curent.items; } } else { previous = current; current = current.next; } } //not found! return null; }

删除节点的方法

2 个答案: