题解 | #链表中倒数最后k个结点#

1. 判断链表是否为空

if (pHead == NULL) return pHead;

2. 采用快慢指针

ListNode* fast = pHead;
ListNode* slow = pHead;

3. 判断k是否大于链表长度

ListNode* dummy = NULL;
while (k--) {
	fast = fast->next;
	if (fast == NULL && k) return dummy;
}

4. 当快指针移动到末尾时，返回慢指针位置

        while (fast) {
            fast = fast->next;
            slow = slow->next;
        }
        return slow;