删除链表的倒数第 N 个结点(链表篇)

给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。

思路：

 

 

代码：

/**
 * 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) {
        ListNode *dummyHead = new ListNode(0);
        dummyHead->next = head;

        ListNode *fast = dummyHead;
        ListNode *slow = dummyHead;

        
        while(n-- && fast->next != NULL)
        {
            fast = fast->next;
        }

        //这里因为删除一个节点，需要找到前一个节点
        //才能进行删除
        fast = fast->next;
        while(fast != NULL)
        {
            fast = fast->next;
            slow = slow->next;
        }

        ListNode *temp = slow->next;

        slow->next = temp->next;

        delete temp;

        return dummyHead->next;

    }
};