19. 删除链表的倒数第 N 个结点

LeetCode链接：19. 删除链表的倒数第 N 个结点 - 力扣（LeetCode）

思路：用快慢指针，fast指针和slow指针相差n个结点，并定义一个temp结点指向head，当fast < n时fast = fast.next，当fast ！= null时，slow = slow.next,fast = fast.next，最后将left.next = left.next.next,并返回ans = temp.next

java代码：

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode temp = new ListNode(0,head);
        ListNode right = head;
        ListNode left = temp;

        for(int i = 0; i < n; i++){
            right = right.next;
        }

        while(right != null){
            left = left.next;
            right = right.next;
        }

        left.next = left.next.next;
        ListNode ans = temp.next;

        return ans;
    }
}