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.
ListNode fast = head, slow = head;
if(head == null){ //链表中两个指针的思想灰常重要，一定要掌握！
return null;
}
while(n != 0){
fast = fast.next;
n--;
}
if(fast == null ){ element is the head //注意特殊情况！！好好想想什么时候要考虑这些边界值
return head.next;
}
while(fast.next != null){
slow = slow.next;
fast = fast.next;
}
slow.next = slow.next.next; //被删的节点不用清除么？为什么=null会有错
