19：Remove Nth Node From End of List【两指针】【链表】

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/*题意：删除链表中从最后一个结点向前的第n个结点 */

/**
 *思路：1、两个指针，first和second，让first指针先走n步
 *      2、接着两个指针同时向后移动，当first指向最后一个
 *         结点时，second->next就是要删除的结点
 *
 */


class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        ListNode *root = new ListNode(0);//root为根结点
        ListNode *second, *first;

        root->next = head;
        first = second = root;

        for(int i = 0; i < n; i ++) { //first指针先走n步
            first = first->next;
        }

        while(first->next != NULL) { //first指针已到链表尾部
            first = first->next;
            second = second->next;
        }
        second->next = second->next->next; //删除second->next结点
        return root->next;
    }
};