Basic linklist operation.

Algorithm:

1. split the linklist
2. reverse the second half
3. merge one by one

Analysis:

1. It is equivalent to finding the last k nodes and make them the head.
2. If k > len(list), k = k % len(list)

Algorithm:

1. let fast be k steps ahead of slow.
2. forward both pointer until fast is the last node
3. now slow is the pre of the last kth node

Algorithm:

1. find (m-1)th node
2. reverse (n-m+1) nodes

Basic linklist operation. It can be done either recursively or iteritively.

Algorithms: while traverse the link, delete cur if cur == cur.next.

The trick is to create two pointers. One is n steps faster.

Normal question, but be aware of the upper bound and lower bound.

1. Use recursion to process node by node.
2. When there is carry, create a new node with val = 1 to add on.

Same as #1, #653. Either use set or two pointers.