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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.
Follow up: Could you do this in one pass?
Example 1:
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| Input: head = [1,2,3,4,5], n = 2
Output: [1,2,3,5]
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Example 2:
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| Input: head = [1], n = 1
Output: []
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Example 3:
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| Input: head = [1,2], n = 1
Output: [1]
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移除串列倒數第n個值。
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| Time: O(n), Space: O(1)
# 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, n: int) -> ListNode:
dummy=ListNode(0)
dummy.next=head
pos_0=pos_n=dummy # ListNode->head 0
for i in range(n): # 0, 1, 2
pos_n=pos_n.next # n=1 n=2
while pos_n.next:
pos_0=pos_0.next # n= 0
pos_n=pos_n.next # n= 2
print(pos_0.val) # n=3
print(pos_n.val) # n=5
pos_0.next=pos_0.next.next
return dummy.next
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