Leetcode 19

Question

Solution

先随便创建一个链表，将其与head链接起来。创建两个指针pre和cur，如果n>0，则n-=1，cur后移。如果n=0，则两个指针一起后移。直到cur.next=None。

# 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:
        ans = pre = cur = ListNode(0)
        pre.next = head
        while cur.next:
            if n:
                n -= 1
            else:
                pre = pre.next
            cur = cur.next
        pre.next = pre.next.next
        return ans.next