如何通过链表进行斐波那契（不是通过递归） [英] How to do fibonacci by linked list ( not by recursive)

问题描述

因为在链表中每个节点只包含一个数字，如


curr-> nodevalue_1 = 0

curr-> nodevalue_2 = 1

sum-> nodevalue = curr-> nodevalue_1 + curr-> nodevalue_2


所以当数字达到> 10或> 100时链接列表需要另一个

节点（我可以将节点添加到前面）但是怎么做呢？


ex：89 + 144 = 233

如何将它添加到总和链表？

请帮忙（记住不是递归问题）

解决方案

" fighterman19" <音响********** @ comcast.net>写了...

因为在链表中每个节点只包含一个数字，如


这是一个要求还是只是你对链表的理解？

curr-> nodevalue_1 = 0
curr-> nodevalue_2 = 1
sum-> nodevalue = curr-> nodevalue_1 + curr-> nodevalue_2

所以当数字上升到> 10或> 100并且链表需要另一个节点时（我可以将节点添加到前面）但是怎么做呢？

ex：89 + 144 = 233
如何将它添加到总和链表？
请帮忙（记住不是递归问题）

有什么问题？你不能用结转来实现添加吗？


0 =>随身携带。

label_1：

next_digit1 + next_digit2 + carry =>结果。

如果结果> 9然后

1 =>随身携带。

结果 - 10 =>结果。

其他

0 =>结账。

结束 - 如果

商店结果

如果have_more_digits转到label_1


我不要这里看不到C ++语言问题，BTW。


Victor

fighterman19写道：

因为在链表中每个节点只包含一个数字，如

curr-> nodevalue_1 = 0
curr-> nodevalue_2 = 1
sum-> nodevalue = curr-> nodevalue_1 + curr-> nodevalue_2

所以当数字达到> 10或> 100且链表需要另一个节点时（我可以添加节点到前面）但怎么办呢？

ex：89 + 144 = 233
如何将它添加到总和链表？
请帮忙（记住不是递归问题）

为什么你会为此使用链表呢？你的问题不是很有意义。


-Kevin

-

我的电子邮件地址有效，但会定期更改。

要联系我，请使用最近发布的地址。

fighterman19写道：

因为在链表中每个节点只包含一个数字，如

curr-> nodevalue_1 = 0
curr-> nodevalue_2 = 1
sum-> nodevalue = curr-> nodevalue_1 + curr-> nodevalue_2

所以当数字达到> 10或> 100且链表需要另一个
时节点（我可以将节点添加到前面）但是怎么做呢？

ex：89 + 144 = 233
如何将它添加到总和链表？

我建议你做一个自己的功课，然后

发布你的代码，而不是要求我们为你做这个？

because in linked list each node contains only one digit like

curr->nodevalue_1= 0
curr->nodevalue_2=1
sum->nodevalue = curr->nodevalue_1 + curr->nodevalue_2

so when the number go up to >10 or >100 and the linked list need another
node (I can add the node to the front ) but how can do the sum ?

ex: 89+144 = 233
how can I add it to the sum linked list?
please help ( remember not a recursive problem)

解决方案

"fighterman19" <fi**********@comcast.net> wrote...

because in linked list each node contains only one digit like
Is that a requirement or just your understanding of linked lists?

curr->nodevalue_1= 0
curr->nodevalue_2=1
sum->nodevalue = curr->nodevalue_1 + curr->nodevalue_2

so when the number go up to >10 or >100 and the linked list need another
node (I can add the node to the front ) but how can do the sum ?

ex: 89+144 = 233
how can I add it to the sum linked list?
please help ( remember not a recursive problem)

What''s the problem? Can''t you implement addition with carry-over?

0 => carry.
label_1:
next_digit1 + next_digit2 + carry => result.
if result > 9 then
1 => carry.
result - 10 => result.
else
0 => carry.
end-if
store result
if have_more_digits go to label_1

I don''t see C++ language issue here, BTW.

Victor

fighterman19 wrote:

because in linked list each node contains only one digit like

curr->nodevalue_1= 0
curr->nodevalue_2=1
sum->nodevalue = curr->nodevalue_1 + curr->nodevalue_2

so when the number go up to >10 or >100 and the linked list need another
node (I can add the node to the front ) but how can do the sum ?

ex: 89+144 = 233
how can I add it to the sum linked list?
please help ( remember not a recursive problem)

Why on earth would you use a linked list for this? Your question doesn''t
make much sense.

-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.

fighterman19 wrote:

because in linked list each node contains only one digit like

curr->nodevalue_1= 0
curr->nodevalue_2=1
sum->nodevalue = curr->nodevalue_1 + curr->nodevalue_2

so when the number go up to >10 or >100 and the linked list need another
node (I can add the node to the front ) but how can do the sum ?

ex: 89+144 = 233
how can I add it to the sum linked list?
please help ( remember not a recursive problem)

May I suggest that you take a stab at doing your own homework, and then
post your code, instead of asking us to do it for you?

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