首先采用了最一般的办法，就是先统计出该链表里面有多少个element，然后再执行一次遍历，遍历到倒数第n个节点的前一个节点，再使用p->next = p->next->next就可以删除倒数第n个节点
但不是one pass实现的，代码如下：
/** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ struct ListNode* removeNthFromEnd(struct ListNode* head, int n){ struct ListNode *dummy = malloc(sizeof(struct ListNode)); struct ListNode *p; int len = 0; int diff = 0; dummy->next = head; //从头开始遍历，找到链表长度 p = head； while(p != NULL) { p = p->next; len++; } //把p指针放回开头dummy处 p = dummy; diff = len - n; while(diff--) { p = p->next; } //删除节点 p->next = p->next->next; return dummy->next; }第二种方法，即one pass：
/** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ struct ListNode* removeNthFromEnd(struct ListNode* head, int n){ struct ListNode *dummy = malloc(sizeof(struct ListNode)); struct ListNode *fast; struct ListNode *slow; dummy->next = head; fast = dummy; slow = dummy; //先把fast往后挪n下 while(n--) { fast = fast->next; } /* 同时挪动fast和slow，直到fast的下一个节点指向NULL，slow就挪到了要删除节点的前一个节点，很巧妙的数学方法，赞 */ while(fast->next != NULL) { fast = fast->next; slow = slow->next; } //删除倒数第n个节点 slow->next = slow->next->next; return dummy->next; }