在恒定时间内将节点插入链表？

如果遇到问题，只需写下来即可：

// First we have a pointer to a node containing element (elm) // with possible a next element.// Graphically drawn as:// p -> [elm] -> ???tmp = new Node();// A new node is created. Variable tmp points to the new node which // currently has no value.// p   -> [elm] -> ???// tmp -> [?]tmp.element = p.element;// The new node now has the same element as the original.// p   -> [elm] -> ???// tmp -> [elm]tmp.next = p.next;// The new node now has the same next node as the original.// p   -> [elm] -> ???// tmp -> [elm] -> ???p.element = y;// The original node now contains the element y.// p   -> [y] -> ???// tmp -> [elm] -> ???p.next = tmp;// The new node is now the next node from the following.// p   -> [y] -> [elm] -> ???// tmp -> [elm] -> ???

您已达到要求的效果，但是它可以提高效率，我敢打赌，您现在就可以了解自己。

写这样的东西更清晰：

tmp = new Node();tmp.element = y;tmp.next = p;p = tmp;

如果p不可变，那当然不起作用。但是，如果p == NULL，则您的算法将失败。

但是我要说的是，如果您对算法有疑问，只需写下效果即可。尤其是对于树和链表，您需要确保所有指针都指向严格的方向，否则会造成混乱。