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coding-interviews

《剑指offer》(第二版)Java实现

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coding-interviews's Issues

链表删除常规解法中可能会存在问题

/**
     * 常规方法,从first开始找到要删除结点的前一个结点,时间复杂度为O(n)
     */
    public void deleteNode_2(Node first, Node toBeDel) {
        if (first == null || toBeDel == null) {
            return;
        }
        // 要删除的就是链表头,它没有前一个结点
        if (first == toBeDel) {
            first = first.next;
        } else {
            Node cur = first;
          	// 找到被删除结点的前一个结点
            while (cur.next != toBeDel) {
                cur = cur.next;
            }
            // cur为toBeDel的前一个结点
            cur.next = cur.next.next;
        }
    }

老哥你好,第 18 题删除链表的节点中,上述代码的 else 语句块的 while循环中,找到被删除结点的前一个结点这个我觉得可能会出现问题,没有中断条件的话,如果要删除的节点不在链表中会一直循环。我觉得应该要加一个中断条件,如果要删除节点不在链表中,并且已经遍历完了链表要退出循环。修改如下:

/**
     * 常规方法,从first开始找到要删除结点的前一个结点,时间复杂度为O(n)
     */
    public void deleteNode_2(Node first, Node toBeDel) {
        if (first == null || toBeDel == null) {
            return;
        }
        // 要删除的就是链表头,它没有前一个结点
        if (first == toBeDel) {
            first = first.next;
        } else {
            Node cur = first;
          	// 找到被删除结点的前一个结点
            while (curr.next != null && cur.next != toBeDel) {
                cur = cur.next;
            }
            if (curr.next == null) {
                return;
            }
            // cur为toBeDel的前一个结点
            cur.next = cur.next.next;
        }
    }

希望老哥看下这样可不可行,谢谢。

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