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;
}