Question

我正在创建一个单独的链接列表，我需要放一个删除方法，但得到错误任何帮助？

public void remove(int index) {
    getNode(index - 1).Next = getNode(index + 1);
}

Answer 1

这里可能发生许多可能的错误。首先，我明确地会在进行任何更改之前进行更多验证，就像

一样

你是第一个元素（那么标题就是第二个元素）;
您是否在最后一个元素上（然后是getNode（index + 1）正确返回NULL或抛出异常）？

Answer 2

正如Apokai正确地指出你在删除时需要格外小心。

以下是三种删除用途的方法：

 //Method to delete the first/head element
    public void deleteAtFirst() {
        if (head == null) {//If list is empty
            System.out.println("Linked List EMPTY!!");

        } else {//else if it is not empty
            head = head.next;//assigning head the address of the next node
        }
    }

    //Method to delete the last element
    public void deleteAtLast() {
        if (head == null) {//If list is empty
            System.out.println("Linked List EMPTY!!");
        } else if (head.next == null) {//else if it has only 2 elements
            head = null;
        } else {//else if it is not empty and has more than 2 elements
            Node tmpNode = head;//taking a temporary node and initialising it with the first/head node
            while (tmpNode.next.next != null) {//looping it till the 'second last' element is reached
                tmpNode = tmpNode.next;
            }
            tmpNode.next = null;//assigning the next address of the second last node as null thus making it the last node
        }
    }

    //Method to delete an element at a given position
    //The first element is situated at 'Position:[0]'
    public void deleteAtPosition(int pos) {
        int size = getSize();//Getting the size of the linked list through a pre-defined method
        if (head == null || pos > size - 1) {//if the position is beyond the scope of current list or the list is empty
            System.out.println("Position: " + pos + "does not exist");
        } else {
            Node prevNode = null;
            Node currNode = head;

            if (pos == size - 1) {//if position is equal to size-1 then essentially the last element is to be deleted
                deleteAtLast();
            } else if (pos == 0) {//if position given is '0' then we need to delete the first element
                deleteAtFirst();
            } else {//else if it is any other valid position
                for (int i = 0; i < pos; i++) {//looping till the desired position
                    prevNode = currNode;//the node just before the required position will be assigned here
                    currNode = currNode.next;//the current node
                }
                prevNode.next = currNode.next;//assigning the next address of previous node to the next of current node thus removing the current node from the list
            }
        }
    }

请注意如果您使用最后一个，则代码中会出现上述两种方法。它还使用另一种方法getSize（）：

//Method to get the size of the list
    public int getSize() {//return type is integer as we are returning 'int size'
        if (head == null) {//if the list is empty
            return 0;
        }
        int size = 1;//initialising size by one
        Node tmpNode = head;//taking a temporary node and initialising it with the first/head node
        while (tmpNode.next != null) {//looping till the end of the list
            tmpNode = tmpNode.next;//incrementing node to the next node 'after every iteration' of the loop
            size++;//incrementing the size
        }
        return size;//returning size
    }

单链表删除值

2 个答案: