## Reorder List

###### DESCRIPTION (credit Leetcode.com)

Given a reference head of type ListNode that is the head of a singly linked list, reorder the list **in-place** such that the nodes are reordered to form the following pattern:

1st node -> last node -> 2nd node -> 2nd to last node -> 3rd node ...

###### EXAMPLES

**Example 1:**
input:

output:

**Example 2:**
input:

output:

## Solution

The first observation is that we can create our final list by merging two smaller lists: the first half of the original list and the reversed second half of the original list.

For example, if our original list is 5 -> 4 -> 3 -> -> 2 -> 1, then the first half is 5 -> 4 -> 3 and the reversed second half is 1 -> 2. We can merge these two lists to get the final list 5 -> 1 -> 4 -> 2 -> 3.

With that established, we can focus on getting the reverse of the 2nd half of the list. This breaks down to first finding the middle node of the list, and then reversing the direction of the nodes starting from the middle node to the end of the list.

Put together, the solution involves three steps, each of which involve 3 fairly common linked list operations:

- Finding the middle of the linked list (using
**fast and slow pointers**). - Reversing the nodes between the middle and the end of the linked list.
- Merging the first half of the linked list with the reversed second half.

#### Step 1: Find the Middle of the Linked List

This step can be done using **fast and slow pointers**, which involves initializing two pointers, fast and slow, at the head of the linked list, and then iterating until fast reaches the end of the list. At each iteration, slow moves one node forward and fast moves two nodes forward. When fast reaches the tail node of the list, slow will point to the middle node in the list.

def reorderList(head):if not head or not head.next:return head# find middle nodeslow = fast = headwhile fast and fast.next:fast = fast.next.nextslow = slow.next# reverse second half of listprev, curr = None, slowwhile curr:next_ = curr.nextcurr.next = prevprev, curr = curr, next_# merge first and reversed second half of listfirst, second = head, prevwhile second.next:first.next, first = second, first.nextsecond.next, second = first, second.nextreturn head

When there are an even number of nodes, fast will equal to None after moving past the tail node, and slow will be the first node of the second half of the list. In the case below, there are 4 nodes, so slow will point to the 3rd node.

def reorderList(head):if not head or not head.next:return head# find middle nodeslow = fast = headwhile fast and fast.next:fast = fast.next.nextslow = slow.next# reverse second half of listprev, curr = None, slowwhile curr:next_ = curr.nextcurr.next = prevprev, curr = curr, next_# merge first and reversed second half of listfirst, second = head, prevwhile second.next:first.next, first = second, first.nextsecond.next, second = first, second.nextreturn head

#### Step 2: Reverse second half of list

At this point, slow points to the middle node in the list. Next, we want to reverse the direction of the pointers of each node starting from slow to the tail of the list.

Reversing the nodes in a linked list is a common problem that can be solved by iterating over each node that needs to be reversed. The key idea is to maintain three pointers, prev, curr, and next_, where prev points to the previous node, curr points to the node with the pointer we want to reverse, and next_ points to the next node in the iteration.

At each iteration, we:

- save the next node in the iteration by setting next_ = curr.next
- reverse the pointer by setting curr.next = prev
- move pointers for the next iteration by set curr = next_ and prev = curr

def reorderList(head):if not head or not head.next:return head# find middle nodeslow = fast = headwhile fast and fast.next:fast = fast.next.nextslow = slow.next# reverse second half of listprev, curr = None, slowwhile curr:next_ = curr.nextcurr.next = prevprev, curr = curr, next_# merge first and reversed second half of listfirst, second = head, prevwhile second.next:first.next, first = second, first.nextsecond.next, second = first, second.nextreturn head

#### Step 3: Merge first half with reversed second half

At this point, when curr is None, then prev will be the head of the reversed second half of the list. We can then merge the first half of the list with the reversed second half by iterating over the two halves and updating the pointers of the nodes.

We can do so by initializing two pointers: first, which points to the head of the first half of the list, and second, which points to the head of the reversed second half of the list, which is initially prev.

def reorderList(head):if not head or not head.next:return head# find middle nodeslow = fast = headwhile fast and fast.next:fast = fast.next.nextslow = slow.next# reverse second half of listprev, curr = None, slowwhile curr:next_ = curr.nextcurr.next = prevprev, curr = curr, next_# merge first and reversed second half of listfirst, second = head, prevwhile second.next:first.next, first = second, first.nextsecond.next, second = first, second.nextreturn head

From there, we want to merge the nodes at first and second together, with first coming before second. To do so, we first set first.next = second. And since we are over-writing first.next, we need to simultaneously advance first = first.next so that we have access to the next node in the first half of the list.

def reorderList(head):if not head or not head.next:return head# find middle nodeslow = fast = headwhile fast and fast.next:fast = fast.next.nextslow = slow.next# reverse second half of listprev, curr = None, slowwhile curr:next_ = curr.nextcurr.next = prevprev, curr = curr, next_# merge first and reversed second half of listfirst, second = head, prevwhile second.next:first.next, first = second, first.nextsecond.next, second = first, second.nextreturn head

Now the first node in the original list and the first node in the reversed second half are connected, so we need to connect the 2nd node in the original list with the first node in the reversed second half. We can do so by setting second.next = first and making sure we simultaneously advance second = second.next so that we have access to the next node in the reversed second half.

def reorderList(head):if not head or not head.next:return head# find middle nodeslow = fast = headwhile fast and fast.next:fast = fast.next.nextslow = slow.next# reverse second half of listprev, curr = None, slowwhile curr:next_ = curr.nextcurr.next = prevprev, curr = curr, next_# merge first and reversed second half of listfirst, second = head, prevwhile second.next:first.next, first = second, first.nextsecond.next, second = first, second.nextreturn head

At this point, the first three nodes in our original linked list are correctly ordered, and first and second are pointing to the next nodes in each half that we need to merge together. So we can continue the merge until second has reached the end of the reversed second half of the list, at which point we can return None.

def reorderList(head):if not head or not head.next:return head# find middle nodeslow = fast = headwhile fast and fast.next:fast = fast.next.nextslow = slow.next# reverse second half of listprev, curr = None, slowwhile curr:next_ = curr.nextcurr.next = prevprev, curr = curr, next_# merge first and reversed second half of listfirst, second = head, prevwhile second.next:first.next, first = second, first.nextsecond.next, second = first, second.nextreturn head

## Code

def reorderList(head):if not head or not head.next:return head# find middle nodeslow = fast = headwhile fast and fast.next:fast = fast.next.nextslow = slow.next# reverse second half of listprev, curr = None, slowwhile curr:next_ = curr.nextcurr.next = prevprev, curr = curr, next_# merge first and reversed second half of listfirst, second = head, prevwhile second.next:first.next, first = second, first.nextsecond.next, second = first, second.nextreturn head

### Edge Cases

#### Empty List

When the linked list is empty, the initial check for head being None will return None immediately.

def reorderList(head):if not head or not head.next:return head# find middle nodeslow = fast = headwhile fast and fast.next:fast = fast.next.nextslow = slow.next# reverse second half of listprev, curr = None, slowwhile curr:next_ = curr.nextcurr.next = prevprev, curr = curr, next_# merge first and reversed second half of listfirst, second = head, prevwhile second.next:first.next, first = second, first.nextsecond.next, second = first, second.nextreturn head

#### Single Node

When the linked list has one node, the initial check for head.next being None will return head immediately.

def reorderList(head):if not head or not head.next:return head# find middle nodeslow = fast = headwhile fast and fast.next:fast = fast.next.nextslow = slow.next# reverse second half of listprev, curr = None, slowwhile curr:next_ = curr.nextcurr.next = prevprev, curr = curr, next_# merge first and reversed second half of listfirst, second = head, prevwhile second.next:first.next, first = second, first.nextsecond.next, second = first, second.nextreturn head

#### Two Nodes

When the linked list has two nodes, the linked list is already in the correct order.

- The while loop to find the middle node iterates once, with slow pointing to the 2nd node.
- The while loop to reversing the second half runs once, but since there is only one node in the 2nd half of the list, no pointers are updated, and prev points to the 2nd node.
- The while loop to merge the two halves doesn't run, as second.next is None.

def reorderList(head):if not head or not head.next:return head# find middle nodeslow = fast = headwhile fast and fast.next:fast = fast.next.nextslow = slow.next# reverse second half of listprev, curr = None, slowwhile curr:next_ = curr.nextcurr.next = prevprev, curr = curr, next_# merge first and reversed second half of listfirst, second = head, prevwhile second.next:first.next, first = second, first.nextsecond.next, second = first, second.nextreturn head

## Complexity Analysis

**Time Complexity:** O(n), where n is the number of nodes in the linked list. We iterate over the linked list three times: once to find the middle node, once to reverse the second half of the list, and once to merge the two halves together. Each iteration takes O(n) time.

**Space Complexity:** O(1). We only use a constant amount of extra space to store pointers and temporary variables. In other words, regardless if the linked list has 10 nodes or 10,000 nodes, we will still use two pointers to find the middle node, three pointers to reverse the second half of the list, and two pointers to merge the two halves together.

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