思路

知道链表的长度
确定要删除的节点的位置
处理特殊情况：
1. 只有一个元素
2. n == count

解题方法

由思路可知

Code

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* removeNthFromEnd(struct ListNode* head, int n) {
    struct ListNode *p = head, *q;
    int count = 0;
    while(p) {
        p = p->next;
        count++;
    }
    q = p = head;
    int postion = 0;
    while(postion < count-n-1 && p->next) {
        p = p->next;
        postion++;
    }
    if(postion == count-n-1) {
        q = p->next;
        p->next = p->next->next;
    } else if(n == count) {
        q = head;
        head = p->next;
    }
    free(q);
    return head;
}