LeetCode 솔루션 분류
[9/28] 19. Remove Nth Node From End of List
본문
Medium
13095548Add to ListShareGiven 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 <= 300 <= Node.val <= 1001 <= n <= sz
Follow up: Could you do this in one pass?
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Runtime: 60 ms, faster than 41.68% of Python3 online submissions for Remove Nth Node From End of List.
Memory Usage: 14 MB, less than 20.71% of Python3 online submissions for Remove Nth Node From End of List.
Memory Usage: 14 MB, less than 20.71% of Python3 online submissions for Remove Nth Node From End of List.
# 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]:
count = 0
curr = head
while curr:
count += 1
curr = curr.next
dummy = ListNode(0, head)
left, right = dummy, head.next
LEFT, RIGHT = count - n, count - n + 2
for i in range(LEFT):
right = right.next
left = left.next
left.next = right
return dummy.next
