Question

我正在尝试实现Java链接列表。每当我尝试删除倒数第二个节点时，程序就会给我一个空指针。

以下是属性：

public class Nodo {
  int num;
  boolean used;
  Nodo next;        
}

到目前为止这是方法：

    public int delete(int val) { // deletes, (1:deleted, 0:not found, -1:empty
                                // list)
    int deleted = 0;
    if (n.used) {
        if (n.next == null) {
            if (n.num == val) {
                n.used = false;
                deleted = 1;
            }
        } else {
            Nodo it = n;
            Nodo itAnt = null;
            while (it.next != null) {
                if (it.num == val) {
                    it.num = it.next.num;
                    it.next = it.next.next;
                    deleted = 1;
                }
                itAnt = it;
                it = it.next;
            }
            if (it.num == val) {
                itAnt.next = null;
                deleted = 1;
            }
        }

    } else {
        deleted = -1;
        System.out.println("Empty list");
    }
    return deleted;
}

我该如何解决这个问题？

Answer 1

当然这段代码会引发异常。这是场景：

it有可能是列表中倒数第二个节点，it.num将等于val。现在，在第一次迭代之后，此while循环将具有以下值：

while (it.next != null) {
    if (it.num == val) {
        it.num = it.next.num;
        it.next = it.next.next;  
        // "it.next" will be null. because "it" was one to last in the list.
        deleted = 1;
    }
    itAnt = it;    // This is the last item in the list
    it = it.next; // Here "it" will be null, because "it.next" is null
}

当第二次迭代时，it.next将抛出NullPointerException，因为it已经为空。

还值得一提的是，您不需要这一行：Nodo it = n;因为它是一个对象引用，n和it都指向同一个对象。因此，无论您在it上做什么，都会直接影响n。

删除单个链表中的任何节点

1 个答案: