203. Remove Linked List Elements【java】

问题原文地址点击打开链接

对链表最基础的处理。代码已写注释。代码如下

    public ListNode removeElements(ListNode head, int val) {
        if (head == null) return head;
        ListNode current = head;//移动的指针
        ListNode pre = null;//用该变量记录上一个节点
        while(current.next != null){
            if (current.val == val){
                //对特殊位置,头指针进行判断
                if (current == head){
                    head = head.next;
                    current = head;
                }else{
                    //当符合条件的点在链表中间部分时
                    ListNode next = current.next;
                    pre.next = next;//将该节点删除,并将该节点的前后两个节点连接
                    current = next;
                }
            }else{
                //移动指针
                pre = current;
                current = current.next;
            }
        }
        if (current.val == val){
            //对末尾位置进行判断
            if (pre == null) return null;
            else pre.next = null;
        }
        return head;
    }

猜你喜欢

转载自blog.csdn.net/amber804105/article/details/80908835
今日推荐