题解 | #快慢指针——删除链表的倒数第n个节点#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param head ListNode类 
     * @param n int整型 
     * @return ListNode类
     */
  /*快慢指针，快指针为游标，慢指针定位要删除的结点*/
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        // write code here
        ListNode *ymm = new ListNode(0);//辅助哑结点方便处理头结点
        ymm->next = head;
        ListNode * slow = ymm;
        ListNode * fast = ymm;
        for(int i =0;i<n-1;i++){
            fast = fast->next;
        }
        while (fast->next!=nullptr) {
            fast = fast->next;
            slow = slow->next;
        }
        ListNode *p = ymm;//辅助p结点找到要要删除的结点前面
        while (p->next!=slow) {
            p = p->next;
        }
        p->next = slow->next;
        return ymm->next;
    }
};