[LeetCode19]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.

给定一个单链表，移除倒数第n个Node，返回链表的head

思路：维护两个指针slow，fast，fast先移动n步，然后slow、fast同时移动，直到fast为空为止，删除slow后面的结点即可

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) {
        ListNode *preHead = new ListNode(0);
        ListNode *fast = preHead, *slow = preHead;
        slow->next = head;
        for(int i = 0; i <= n; ++i)
            fast = fast->next;
        while(fast)
        {
            fast = fast->next;
            slow = slow->next;
        }
        slow->next = slow->next->next;
        return preHead->next;
    }
};