面试题 02.01. 移除重复节点

链接

面试题 02.01. 移除重复节点

思路

记录已经出现过的结点，添加未出现过的结点，三种方法都在代码里，并不难理解

代码

public ListNode removeDuplicateNodes(ListNode head)
    {
        //传入为空或只有一个结点，则返回head
        if (head == null || head.next == null)
        {
            return head;
        }

        ListNode[] nodes = new ListNode[20001];
        nodes[head.val] = head;
        ListNode cur = head;
        while (head.next != null)
        {
            ListNode temp = head.next;
            //没有该元素
            if (nodes[temp.val] == null)
            {
                nodes[temp.val] = temp;
                head = head.next;
            } else
            {
                head.next = head.next.next;
            }
        }
        head.next = null;
        return cur;
    }

    public ListNode removeDuplicateNodes2(ListNode head)
    {
        //传入为空或只有一个结点，则返回head
        if (head == null || head.next == null)
        {
            return head;
        }

        ListNode[] nodes = new ListNode[20001];
        nodes[head.val] = head;


        ListNode cur = head.next;
        ListNode prev = head;
        while (cur != null)
        {
            //没有该元素
            if (nodes[cur.val] == null)
            {
                nodes[cur.val] = cur;
                prev.next = cur;
                prev = prev.next;
            }
            cur = cur.next;
        }
        prev.next = null;
        return head;
    }

    public ListNode removeDuplicateNodes1(ListNode head)
    {
        //传入为空或只有一个结点，则返回head
        if (head == null || head.next == null)
        {
            return head;
        }
        Map<Integer, Integer> map = new HashMap<>();
        map.put(head.val, 1);

        ListNode cur = head.next;
        ListNode prev = head;
        while (cur != null)
        {
            //如果没有该数字则加入 map
            if (!map.containsKey(cur.val))
            {
                map.put(cur.val, 1);
                prev.next = cur;
                prev = prev.next;
            }
            cur = cur.next;
        }
        prev.next = null;
        return head;
    }