19. Remove Nth Node from End of List，删除倒数第N个节点

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]

# 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]:
        nexts = []
        p = head
        while p:
            nexts.append(p)
            p = p.next
        if len(nexts) == n:
            return head.next
        nexts[-n-1].next = nexts[-n].next
        return head

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        vector<ListNode*> arrNode;
        arrNode.reserve(64);
        ListNode *p { head };
        while(p){
            arrNode.push_back(p);
            p = p->next;
        }
        size_t s { arrNode.size() } ;
        if(n == s)
            return head->next;
        arrNode[s-n-1]->next = arrNode[s-n]->next;
        return head;
    }
};

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* removeNthFromEnd(struct ListNode* head, int n){
    int size = 0;
    struct ListNode *p = head;
    while(p){
        ++size;
        p = p->next;
    }
    if(size == n)
        return head->next;
    int move = size-n-1;
    p = head;
    while(move != 0){
        p = p->next;
        --move;
    }
    p->next = p->next->next;
    return head;
}