C++解决输出链表中倒数k个结点的问题

class Solution {
public:
    
    Listnode* FindKthToTail(ListNode* pHead, int k) {
        // 空链表直接返回
        if(pHead == NULL)
            return pHead;
        // 快慢指针
        ListNode *fast = pHead;
        ListNode *slow = pHead;
        // 让快指针先走k步
        while(k--)
        {
            if(fast == NULL)
                return NULL;
            fast = fast->next;
        }
        // 当快指针走完时，慢指针的位置就是倒数第k个结点
        while(fast != NULL)
        {
            fast = fast->next;
            slow = slow->next;
        }
        return slow;
    }
};

class Solution {
public:
    
    ListNode* FindKthToTail(ListNode* pHead, int k) {
        // write code here
        ListNode* p=pHead;
        ListNode* q=pHead;
        if(pHead==NULL)return NULL;
        while(k--)
        {
            if(q==NULL)return NULL;
            q=q->next;
            
            //k--;
        }
        //if(q==NULL)return pHead;
        while(q)
        {
            p=p->next;
            q=q->next;
        }
        return p;
    }
};