LeetCode/19_Remove_Nth_Node_From_End_of_List.py at main · AnmolPatel20/LeetCode · GitHub

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"""
Given the head of a linked list, remove the nth node from the end of the list and return its head.


Example 1:


Input: head = [1,2,3,4,5], n = 2
Output: [1,2,3,5]
Example 2:

Input: head = [1], n = 1
Output: []
Example 3:

Input: head = [1,2], n = 1
Output: [1]


Constraints:

The number of nodes in the list is sz.
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz


Follow up: Could you do this in one pass?

"""

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
        dummy = ListNode(0)
        dummy.next = head

        firstPtr = dummy
        secondPtr = dummy

        for _ in range (n+1):
            secondPtr = secondPtr.next

        while secondPtr:
            firstPtr = firstPtr.next
            secondPtr = secondPtr.next

        firstPtr.next = firstPtr.next.next

        return dummy.next