LeetCode:Remove Nth Node From End of List

题目描写叙述：

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.

思路：设置两个指针first跟second。first指针先移动n步。若此时first指针为空，则表示要删除的是头节点，此时直接返回head->next就可以。

假设first指针不为空，则将两个指针一起移动，直到first指针指向最后一个节点，令second->next=second->next->next就可以删除第你n个节点。

代码：

ListNode *removeNthFromEnd(ListNode *head, int n) {
    if(head == NULL || head->next == NULL)
        return NULL;
    ListNode * first = head;
    ListNode * second = head;
    for(int i = 0;i < n;i++)
        first = first->next;
    if(first == NULL)
        return head->next;
    while(first->next != NULL)
    {
        first = first->next;
        second = second->next;
    }
    second->next = second->next->next;
    return head;
    }