Remove Nth Node From End of List（easy）

【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.

【题意】

  删除链表的倒数第n个元素

【分析】

    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即可。如果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即可删除第你n个节点。

【实现】

public ListNode removeNthFromEnd(ListNode head, int n) {    if(head == null)        return null;     ListNode fast = head;    ListNode slow = head;     for(int i=0; i<n; i++){        fast = fast.next;    }     //if remove the first node    if(fast == null){        head = head.next;        return head;    }     while(fast.next != null){        fast = fast.next;        slow = slow.next;    }     slow.next = slow.next.next;     return head;}

c++ 

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;    }