代码题（67）— 删除链表的倒数第 N 个结点、链表中倒数第 K个节点

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if(head == nullptr || n<=0)
            return nullptr;
        ListNode* dummy = new ListNode(-1);
        dummy->next = head;
        ListNode* fast = dummy;
        ListNode* slow = dummy;
        for(int i=0;i < n+1 ;++i){
            if(fast == nullptr){
                return nullptr;
            }
            fast = fast->next;
        }
        while(fast) {
            fast = fast->next;
            slow = slow->next;
        }
        slow->next = slow->next->next;
        return dummy->next;
    }
};

   ListNode* FindKthToTail(ListNode* head, int k) {
        if(head == nullptr || k <= 0){
            return nullptr;
        }
        ListNode* fast = head, * slow = head;
        for(int i=0;i<k;++i){
            if(fast == nullptr)
                return nullptr;
            fast = fast->next;
        }
        while(fast){
            fast = fast->next;
            slow = slow->next;
        }
        return slow;
    }