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

给定一个链表，删除链表的倒数第 n 个节点，并且返回链表的头结点。

整体思路是让前面的指针先移动n步，之后前后指针共同移动直到前面的指针到尾部为止
代码

/**
 * 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;
        //让start结点先走N步
        while(n != 0){
            start = start.next;
            n--;
        }
        //start和end 同时走
        while(start.next != null){
            start = start.next;
            end = end.next;
        }
        end.next = end.next.next;
        return pre.next;
    }
}