链表中node.next指向含义及删除节点代码逻辑疑问

Let's walk through your questions step by step, using your code and diagram context. First, here's the deletion code you shared for reference:

if(position ==1) { 
    ListNode temp = head; 
    head = head.next; 
    temp.next = null; 
    return temp; 
}else { 
    ListNode previous = head; 
    int count =1; 
    while(count < position -1) { 
        previous = previous.next; 
        count++; 
    } 
    ListNode current = previous.next; 
    previous.next = current.next; 
    current.next = null; 
    return current; 
}

1. What does head.next point to in your diagram? #

In the diagram ｜10｜＿｜--> | 8 | __ |, let's clarify how LinkedList pointers work:

• The head variable holds a reference to the entire node containing 10 (the first node in the list).
• head.next is the next property of that first node, which stores a direct reference to the entire node containing 8 (the second node).

There's no "next node space" — each node is a discrete object, and the next pointer simply points to the next node object in the sequence.

2. Purpose of current.next = null; & your confusion about previous.next #

Let's break this down:

• What does current.next = null; do?
  current is the node you're deleting (the position 3 node with value 15). Setting its next to null breaks the link between this node and the rest of the list. In garbage-collected languages (like Java, which this code appears to be written in), this ensures the deleted node isn't holding onto references to other nodes, making it eligible for garbage collection (prevents memory leaks). Even if you skip this step, the node would be removed from the list, but it's a clean, standard practice to tidy up references this way.

• Is previous.next = current.next wrong?
  No — this line is completely correct! Let's map it to your list structure:

  • previous is the node at position 2 (value 8).
  • current is the node at position 3 (value 15), so current.next points directly to the node at position 4 (value 11).
  • Setting previous.next = current.next makes the position 2 node point directly to the position 4 node, effectively removing the position 3 node from the list. If you changed it to previous.next = current.next.next, you'd skip the position 4 node entirely (pointing to whatever comes after 11), which is not what you want for a standard deletion.

3. Verifying head.next & head.next.next in your updated diagram #

Assuming your updated diagram shows:

• head pointing to the first node (value 10)
• head.next pointing to the second node (value 8)
• head.next.next pointing to the third node (value 15)

This is 100% correct. Each .next traverses one node forward in the list:

• head = first node
• head.next = second node
• head.next.next = third node
• And so on.

内容的提问来源于stack exchange，提问作者吳芯緰