feat(work): add solution for LeetCode 19 Remove Nth Node From End of List

work/1/Medium/Linked List/19.remove-nth-node-from-end-of-list.py

@@ -0,0 +1,69 @@
+"""
+19. Remove Nth Node From End of List
+Difficulty: Medium
+https://leetcode.com/problems/remove-nth-node-from-end-of-list/
+
+──────────────────────────────────────────────────
+
+Given the head of a linked list, remove the n^th 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: ListNode | None, n: int) -> ListNode | None:
+        dummy = ListNode(0)
+        dummy.next = head
+
+        fast = dummy
+        for _ in range(n):
+            if fast.next is None:
+                return head
+            fast = fast.next
+
+        slow = dummy
+        while fast.next is not None:
+            slow = slow.next
+            fast = fast.next
+
+        slow.next = slow.next.next
+
+        return dummy.next