一个单链表问题
How to write code to delete a specific node in a single link list （单链表） that takes O(1) time? That is, the time deleting a node is the same (independent from the length of the list.) Link list uses pointers,
not hash. Input is a pointer of the deleting node. Show your algorithm with pseudo code. Hint: just 3 steps.
我回复了 q = p;
p = p-> next;
free(q);
他又回复说can you rework #7? a node in single link list has two fields: data, pointer (pointing to the next node).
our answer "p = p-> next; " should be more specific.
请高手帮忙.
问题的关键是不是怎么找出指向p指针的前一个节点的指针?
[解决办法]
就是用p的下一个节点(p-> next())的覆盖当前要删除的节点的data并完成链表的连接,然后删除当前节点p的下一个节点.
即:
q = p-> next();
p-> data = q-> data;
p-> next() = q-> next();
delete q;
