题解 | #删除链表的倒数第n个节点#

//快慢指针难点在于while (fast)判断是fast到达最后一个，然后就是slow的起始位置选取，如果都从pHead出发
//则slow会到达倒数k个位置上，无法删除节点，于是new个节点，让slow和fast相距k+1个距离，避免越界


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 pHead ListNode类
     * @param k int整型
     * @return ListNode类
     */
    ListNode* removeNthFromEnd(ListNode* pHead, int k) {
        ListNode* nHead = new ListNode(-1);
        nHead->next = pHead;
        ListNode* slow = nHead;
        ListNode* fast = pHead;
        while (k) {
            fast = fast->next;
            cout << k << " ";
            k--;
        }
        while (fast) {
            fast = fast->next;
            slow = slow->next;

        }
        slow->next = slow->next->next;

        return nHead->next;
    }
};