1. Jan 30, 2017

### FallenApple

So I have a linked list a->b->c->d

and I want to make another list

b->a->d->c

in a systematic way by interchanging pointers. But for some reason it isn't working. All I did was move the link between b and c into b to a. Then the link from a to b, changed to a to d. Then the one connecting c to d is reversed. I'm not sure what went wrong.

Code (Python):

class ListNode(object):
def __init__(self, x):
self.val = x
self.next = None

A=['a','b','c','d']
#filling the linked list with abcd
for i in A[1:]:
temp.next=ListNode(i)
temp=temp.next

while(point):
print(point.val)
point=point.next

print("   ")

#attempting to alter the list

while(point):
print(point.val)
point=point.next

The output is

a
b
c
d

b
a
c
d

2. Jan 30, 2017

### rcgldr

You have to avoid using any links that you've changed, either with careful ordering of operations and/or using temp pointers to avoid conflicts.

For example, if swapping two nodes in a linked list, if the nodes are adjacent, the next pointers are effectively rotated, but if the nodes are not adjacent, then two pairs of next pointers are swapped. This can be handled with common code by first swapping the external pointers to the two nodes, followed by swapping the nodes internal next pointers.

Given a list a->b->c->d->e->f, to swap b and e, swap(a->next, d->next), swap(b->next, e->next). To swap c and d, swap(b->next, c->next), swap(c->next, d->next).

3. Feb 8, 2017

### Inventive

Implement as a doubly linked list. In that data structure any given node knows who's it is attached to. Be very careful with the pointer manipulation or orphaned nodes (and thus memory leaks) will occur

4. Feb 8, 2017

### Inventive

Also make sure to test the link pointers for null so that the node is valid. Usually the "next" pointer is null to indicate there are no further nodes