leetcode_gym/_problems/reorder_list.md

layout, title, last_solved, category, leetcode_url, status

layout	title	last_solved	category	leetcode_url	status
page	Reorder List		linked list	https://leetcode.com/problems/reorder-list/	Attempted

Problem

Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You may not modify the values in the list's nodes, only nodes itself may be changed.

Example 1:

Given 1->2->3->4, reorder it to 1->4->2->3.
Example 2:

Given 1->2->3->4->5, reorder it to 1->5->2->4->3.

Solution

With Stack:

• make a queue(), push all nodes to it
• pop off left and right side of q, alternating each time
• as you pop off, rebuild linked list (make sure to set node.next to None)

{% highlight python %}

def reorderList(self, head: ListNode) -> None:

    # push all into list
    q = deque()
    while head:
        q.append(head)
        head = head.next
    
    
    dummy = ListNode(0)
    cur = dummy
    even = False
    
    while q:
        
        node = q.pop() if even else q.popleft()
        node.next = None
        
        cur.next = node
        cur = cur.next
        
        even ^= True
    
    return dummy.next

{% endhighlight %}

---

Without Stack

• split list in half
• reverse second half
• merge lists

{% highlight python %}

def reorderList(self, head: ListNode) -> None:

    if not head: return None
    
    # 1 2 3 4
    # 8 7 6 5

    
    # 1) split list in half
    prev, slow, fast = None, head, head
    
    
    
    while fast and fast.next:
        
        prev = slow     # prev is "tail" of slow
        slow = slow.next
        fast = fast.next.next
    
    # print(prev.val, slow.val, fast)
    
    # current state
    # 1 2 3 4
    #   p s   f
    
    
    prev.next = None
    
    l1 = head
    
    # 1 2
    # l p
    #
    # 3 4
    # s f
    
    # 2) reverse second list
    def reverse(head):
        
        prev, curr, next = None, head, None
        
        while curr:
            next = curr.next
            curr.next = prev
            prev = curr
            curr = next
        
        return prev
    
    
    l1 = head
    l2 = reverse(slow)
    print(l1.val, l2.val)
    
    # 1 2
    # 4 3
    
    # 3) time to merge
    dummy = ListNode(0)
    curr = dummy
    
    while l1 and l2:
        
        if l1:
            curr.next = l1
            l1 = l1.next
            curr = curr.next
        if l2:
            curr.next = l2
            l2 = l2.next
            curr = curr.next
    
    if l1:
        curr.next = l1
    
    if l2:
        curr.next = l2
    
    return dummy.next

{% endhighlight %}