【LeetCode】19. Remove Nth Node From End of List

题目：

Given a linked list, remove the nth node from the end of list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.

   After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

提示：

此题可以通过双指针（两个指针之间相差n个间隔），这样可以保证一次遍历链表就完成题目中的要求。

有一些需要注意的地方：

• 被删去的节点记得要释放资源；
• 可以创建一个指向head的节点，然后初始状态下双指针均指向创建的这个节点，这样做可以让算法可以处理删除head节点的情况。

代码：

/**
 * 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) 
    {
        if (!head) return nullptr;

        ListNode new_head(-1);
        new_head.next = head;

        ListNode *slow = &new_head, *fast = &new_head;

        for (int i = 0; i < n; i++)
            fast = fast->next;

        while (fast->next) 
        {
            fast = fast->next;
            slow = slow->next;
        }

        ListNode *to_be_deleted = slow->next;
        slow->next = slow->next->next;

        delete to_be_deleted;

        return new_head.next;
    }
};