[leetcode] 19. Remove Nth Node From End of List

题目

Given 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 <= 30
• 0 <= Node.val <= 100
• 1 <= n <= sz

Follow up: Could you do this in one pass?

思路

快慢指针，块指针事先步进n格，随后与慢指针一起步进，直至快指针到达链表尾部，此时使用慢指针进行删除元素的操作。注意有可能需要删掉头节点，因此最好使用一个pre_head节点指向头节点。

代码

python版本：

# 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]:
        l=r=root=ListNode(0,head)
        for _ in range(n):
            r=r.next
        while r.next:
            l=l.next
            r=r.next
        l.next=l.next.next
        return root.next