19.删除链表的倒数第N个结点

输入：head = [1,2,3,4,5], n = 2
输出：[1,2,3,5]

题解：

双指针，第一个指针和第二个指针之间相差N个位置，当第一个指针指到空指针时，第二个指针指向的位置就是要删除的第N个结点的前面的指针。

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        int count =0;
        ListNode *abandant = new ListNode(0,head);
        ListNode *flag,*pre_flag= abandon;
        flag = head;
        while(flag){
            count ++;
            if (count>n){
                pre_flag = pre_flag->next;
            }
            flag = flag->next;
        }
        pre_flag->next = pre_flag->next->next;
        ListNode* ans = abandon->next;
        delete abandon;

        return ans;
    }
};

class Solution(object):
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        abandon = ListNode (0,head)
        flag = head
        pre_flag = abandon

        while flag:
            if n<=0:
                pre_flag = pre_flag.next
            flag = flag.next
            n -= 1
        pre_flag.next = pre_flag.next.next
        return abandon.next