删除链表的倒数第N个节点

题目链接：https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list/
题目描述：

题解：
方法1：先遍历链表长度，再查找倒数第n个节点位置。时间复杂度O(L)

/**
 * 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* dummy = new ListNode();
        dummy->next = head;
        ListNode* p = dummy;
        int length = 0;
        while(p->next != nullptr)
        {
            p = p->next;
            length++;
        }
        p = dummy;
        int k = length - n;
        while(k != 0)
        {
            p = p->next;
            k--;
        }
        p->next = p->next->next;
        return dummy->next;
    }
};

方法2：双指针法，第一个指针比第二个指针先n个节点遍历，当第一个指针位于链表尾时，第二个指针正好位于倒数第n个节点处。为了删除方便，第二个节点从哑节点开始遍历，第一个节点从头节点开始遍历。

/**
 * 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* dummy = new ListNode();
        dummy->next = head;
        ListNode* first = head;
        ListNode* second = dummy;
        for(int i = 0; i < n; i++)
        {
            first = first->next;
        }
        while(first != nullptr)
        {
            first = first->next;
            second = second->next;
        }
        second->next = second->next->next;
        return dummy->next;
    }
};