Given a linked list, remove the n-th node from the end of list and return its head.
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.
Given n will always be valid.
Could you do this in one pass?
1. Fast and slow pointers
# Time: O(n) # Space: O(1) # Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode: fast, slow = head, head for _ in range(n): # if not fast.next: # return None fast = fast.next while fast and fast.next: fast = fast.next slow = slow.next if not fast: return slow.next else: slow.next = slow.next.next return head # 208/208 cases passed (32 ms) # Your runtime beats 58.65 % of python3 submissions # Your memory usage beats 100 % of python3 submissions (12.7 MB)