java - 无法从链表中删除元素？

Question

我只是在练习我的数据结构，并尝试制定一种从单链表中删除重复项的方法。这就是我所拥有的:

        void removeDup() {
            Node temp = head;
            Node cur = null;
            String s = "";

            while(temp!=null) {
                cur = temp;

                if(!s.contains(temp.data + "")) {
                    s += temp.data + "";
                }
                else {
                    cur.next = temp.next;
                }

                temp = temp.next;
            }
        }

执行此方法后打印链表没有显示任何变化。我相信这是因为我没有正确地将前一个链接链接到当前链接的.next 值，但对我来说一切看起来都是正确的。我对其进行调试，似乎正确删除了该节点，但当我随后打印出链表时，重复的节点仍然出现。有建议吗？

Answer 1

代码复制自https://www.geeksforgeeks.org/remove-duplicates-from-an-unsorted-linked-list/ :

方法1 - 暴力破解，找到所有的两个节点对，看看它们是否有相同的值，不确定调用System.gc()是否是一个好主意:

/* Function to remove duplicates from an 
   unsorted linked list */
void remove_duplicates() { 
    Node ptr1 = null, ptr2 = null, dup = null; 
    ptr1 = head; 

    /* Pick elements one by one */
    while (ptr1 != null && ptr1.next != null) { 
        ptr2 = ptr1; 

        /* Compare the picked element with rest 
            of the elements */
        while (ptr2.next != null) { 

            /* If duplicate then delete it */
            if (ptr1.data == ptr2.next.data) { 

                /* sequence of steps is important here */
                dup = ptr2.next; 
                ptr2.next = ptr2.next.next; 
                System.gc(); 
            } else /* This is tricky */ { 
                ptr2 = ptr2.next; 
            } 
        } 
        ptr1 = ptr1.next; 
    } 
}

方法2 - 使用hashset来帮助检测重复，我个人更喜欢这种方法:

 /* Function to remove duplicates from a 
       unsorted linked list */
    static void removeDuplicate(node head)  
    { 
        // Hash to store seen values, changed a little to compile for Java 8
        HashSet<Integer> hs = new HashSet<Integer>(); 

        /* Pick elements one by one */
        node current = head; 
        node prev = null; 
        while (current != null)  
        { 
            int curval = current.val; 

             // If current value is seen before 
            if (hs.contains(curval)) { 
                prev.next = current.next; 
            } else { 
                hs.add(curval); 
                prev = current; 
            } 
            current = current.next; 
        } 

    }