19. 删除链表的倒数第N个节点

19. 删除链表的倒数第N个节点

题解

预设指针pre，start，end，使pre.next = head，start比end多走n步，n—，之后找到要删除的结点的上一个节点（技巧是start.next == null跳出循环），删除即可。最后返回pre.next，因为head结点可能已经被删除。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
    ListNode pre = new ListNode(0);
                pre.next = head;
                ListNode start = pre, end = pre;
                while(n != 0) {
                    start = start.next;
                    n--;
                }
                while(start.next != null) {
                    start = start.next;
                    end = end.next;
                }
                end.next = end.next.next;
                return pre.next;
    }
}