leetcode_gym/_problems/removeNthNodeFromEndOfList.md

layout, title, last_solved, category, leetcode_url, status

layout	title	last_solved	category	leetcode_url	status
page	Remove Nth Node from End of List	2020-06-06	linked list	https://leetcode.com/problems/remove-nth-node-from-end-of-list	Solved

Problem

Given a linked list, remove the n-th node from the end of list and return its head.

Example:

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

Solution

{% highlight python %}

def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:

if not head: return None
if head.next is None: return None

slow = head
fast = head

# move fast up n times
for _ in range(n):
    fast = fast.next


# This situation would happen when we are required to del the first node (n = len(List))
# Also, it can handle the [] case
if not fast:
    return slow.next

# move fast to the last node
while fast.next:
    slow = slow.next
    fast = fast.next
    
# do removal
slow.next = slow.next.next

return head

{% endhighlight %}

---

With dummy node. Dummy node helps with edge case where you want to remove the first element.

{% highlight python %}

def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:

dummy = ListNode(0)
dummy.next = head

slow, fast = dummy, dummy

# advance fast pointer so gap between slow and fast are n apart
for _ in range(n+1):
    fast = fast.next


# advance both pointers, maintain gap
while fast:
    slow = slow.next
    fast = fast.next


# remove the nth node
slow.next = slow.next.next

return dummy.next

{% endhighlight %}