https://leetcode.com/problems/remove-nth-node-from-end-of-list/

使用双指针法，可以仅遍历一次完成节点的定位

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* removeNthFromEnd(ListNode* head, int n) {        ListNode* l = new ListNode(0);        l = head;        ListNode* r = new ListNode(0);        r = l;        while(n--)            r = r->next;        //用于判断head是否为单节点链表        if(r == NULL)            return head->next;        while(r != NULL && r->next != NULL)        {            r = r->next;            l = l->next;        }        ListNode* temp = new ListNode(0);        temp = l->next;        l->next = l->next->next;        if(temp)            delete(temp);        return head;    }};