LeetCode:Remove Nth Node From End of List

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.

这个题很早就做过了，不难，不过小细节很多，最后还是调试通过的，汗！
注意点：用一个变量记录node个数，当n==number of nodes,时删除head

	public ListNode removeNthFromEnd(ListNode head, int n) {
	if(head==null) return head;
	ListNode firstNode=head, prv=head, secNode=head;
	if(n==1 && head.next == null) return null;
	int i = 0, num=0;
	while(firstNode!=null){
	num++;
	firstNode = firstNode.next;
	if(i>=n){
	prv = secNode;
	secNode = secNode.next;
	}
	i++;
	}
	if(num==n)
	head = head.next;
	else
	prv.next = secNode.next;
	return head;
	}

view raw gistfile1.java hosted with ❤ by GitHub