public ListNode reverseList(ListNode head) {
    ListNode prev = null;
    ListNode current = head;
    ListNode next = null;
    
    while (current != null) {
        next = current.next;  // 保存下一个节点
        current.next = prev;  // 反转当前节点的指针
        prev = current;       // 移动prev指针
        current = next;       // 移动current指针
    }
    
    return prev;  // prev现在是新的头节点
}

def reverseList(head):
    if not head or not head.next:
        return head
    
    reversed_head = reverseList(head.next)
    head.next.next = head  # 反转指针方向
    head.next = None       # 断开原指针
    
    return reversed_head

// 错误的反转实现
function reverseList(head) {
    let prev = null;
    let current = head;
    
    while (current) {
        current.next = prev;  // 直接修改next指针，丢失了原next节点
        prev = current;
        current = current.next;  // current已经是prev了，导致无限循环
    }
    
    return prev;
}

// 未考虑空链表情况
struct ListNode* reverseList(struct ListNode* head) {
    struct ListNode* prev = NULL;
    struct ListNode* curr = head;
    struct ListNode* next = NULL;
    
    while (curr != NULL) {
        next = curr->next;
        curr->next = prev;
        prev = curr;
        curr = next;
    }
    
    return prev;  // 如果head为NULL，返回NULL
}