删除链表的倒数第n个结点

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {

        //链表长度为5(len),倒数第2（n）个节点，正数第4(len - n + 1)个节点
        //考虑删除第一个节点的特殊情况，则添加一个头结点。
        ListNode h = new ListNode(0);
        h.next = head;
        int len = 0;       
        while(head != null){
            len++;
            head = head.next;
        }
        int count = len - n + 1;
        ListNode cur = h;
        int i = 1;
        while(i < count){
            cur = cur.next;
            i++;
        }
        cur.next = cur.next.next;
        return h.next;
    }
}

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {

        //要删除的节点是倒数第n个，通过快慢指针的方法
        ListNode h = new ListNode(0);
        h.next = head;
        //已有前提保证n是有效的
        //快指针先走n个位置
        ListNode low = h, high = h;
        int i = 0;
        while(i < n){
            high = high.next;
            i++;
        }
        //此时快慢指针再一起出发
        while(high.next != null){
            low = low.next;
            high = high.next;
        }
        low.next = low.next.next;
        return h.next;

    }
}