删除链表的倒倒数第k个节点

思路

• 删除倒数 第k个，即倒数第k + 1个的后继指向倒数第k -1个
• 所以要找到倒数第k + 1个

虚拟头节点出发，快慢指针

• 快的先走n部
• 然后快慢一起走，直到快的最到最后一个while(fast && fast ->next)

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
       
        auto dummy = new ListNode(-1);
        dummy -> next = head;
        
        auto fast = dummy;
        while(n -- ) fast = fast -> next;
        
        auto slow = dummy;
        while(fast && fast ->next)
        {
            fast = fast -> next;
            slow = slow -> next;
        }
        
        slow -> next = slow -> next -> next;
        
        return dummy -> next;
    }
};