copy link with random additional pointers# JobHunting - 待字闺中
c*2
1 楼
复制linked list。 已知每个节点有两个pointer，一个指向后一个节点，另一个指向
其他任意一节点。 O(n)时间内，无附加内存，复制该linked list。（存储不连续）
其他任意一节点。 O(n)时间内，无附加内存，复制该linked list。（存储不连续）
s*e
2 楼
I saw this problem earlier here. The answer is:
(imagine you have a->b->c->d, lets call the other pointer as "other"
1. duplicate each node and insert after itself. Now you get a->a->b->b->c->c->d->d
2. node n=head
while(n!=null)
n->next->other = n->other->next
n=n->next->next
3. separate the two lists
head2 = head->next
node n=head
while(n->next->next !=null)
n->next->next = n->next->next->next
n->next = n->next->next
n=n->next
n->next = null
n->next->next = null
不知道implement有没有错，不过idea就是这样
(imagine you have a->b->c->d, lets call the other pointer as "other"
1. duplicate each node and insert after itself. Now you get a->a->b->b->c->c->d->d
2. node n=head
while(n!=null)
n->next->other = n->other->next
n=n->next->next
3. separate the two lists
head2 = head->next
node n=head
while(n->next->next !=null)
n->next->next = n->next->next->next
n->next = n->next->next
n=n->next
n->next = null
n->next->next = null
不知道implement有没有错，不过idea就是这样
c*2
3 楼
very good idea.
1) First pass: duplicate each node
2) Second pass: populate new other pointers
3) Third pass: split the list
Thanks!
1) First pass: duplicate each node
2) Second pass: populate new other pointers
3) Third pass: split the list
Thanks!
h*d
4 楼
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