Remove Nth Node From End of List

Remove Nth Node From End of List

Given a linked list, remove the n^th node from the end of list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.

   After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

Idea: use a ring buffer

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        ListNode **buffer = new ListNode*[n + 1];
        memset(buffer, NULL, (n + 1) * sizeof(ListNode*));
        int ptr = 0;
        ListNode *current = head;
        
        while (current) {
            buffer[ptr] = current;
            ptr = (ptr + 1) % (n + 1);
            current = current->next;
        }
        
        if (buffer[ptr] == NULL) {
            head = n > 1 ? buffer[1] : NULL;
        }
        else {
            ListNode *last = buffer[ptr];
            last->next = last->next->next;
        }
        delete[] buffer;
        return head;
    }
};