算法: 移除单链表的倒数第N个节点 19. Remove Nth Node From End of List

19. Remove Nth Node From End of List

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?

1. 计算总长度，减去N，即可得到顺序的需要跳过的节点

# 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]:
        dummy = ListNode(0)
        dummy.next = head
        first = dummy
        l = 0
        while first:
            l += 1
            first = first.next
        l -= n
        first = dummy
        while l > 1:
            l -= 1
            first = first.next

        first.next = first.next.next

        return dummy.next

2. 双指针解法，保持N的距离

# 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]:
        dummy = ListNode(0)
        dummy.next = head
        first, second = dummy, dummy
        for i in range(n+1):
            first = first.next
        
        while first:
            first = first.next
            second = second.next
        
        second.next = second.next.next

        return dummy.next