题解 | #删除链表的倒数第n个节点#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head ListNode类 
     * @param n int整型 
     * @return ListNode类
     */
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        // write code here
        ListNode* dummy = new ListNode(-1);
        dummy->next = head;
        ListNode* lnode = dummy->next;
        ListNode* rnode = dummy->next;
        for(int i = 0; i < n; i ++){
            if(rnode != NULL){
                rnode = rnode->next;
            }else{
                return lnode = NULL;
            }
        }
        ListNode* prelNode = dummy; //保存lnode前一个，也就是倒数第n+1节点
        while(rnode){
            prelNode = prelNode->next;
            lnode = lnode->next;
            rnode = rnode->next;
        }
        prelNode->next = lnode->next;
        return dummy->next;
    }
};