【链表】- 删除链表的中间节点和a/b处的节点

【题目】

给定链表的头节点head，实现删除链表的中间节点的函数

给定链表的头节点head，整数a和b，实现删除位于a/b处节点的函数

//删除链表的中间节点
public class Node{
	public int value;
	public Node next;

	public Node(int data){
		this.value = data;
	}
}

public Node removeMidNode(Node head){
	if(head == null || head.next == null){
		return head;
	}

	if(head.next.next == null){
		return head.next;
	}
	Node pre = head;
	Node cur = head.next.next;
	while(cur.next != null && cur.next.next != null){
		pre = pre.next;
		cur = cur.next.next;
	}
	pre.next = pre.next.next;
	return head;
}

//删除链表的a/b处的节点
public Node removeByration(Node head, int a, int b){
	if(a < 1 || a > b){
		return head;
	}
	int n = 0;
	Node cur = head;
	while(cur != null){
		n++;
		cur = cur.next;
	}
	n = (int)Math.ceil(((doule)(a*n))/(double)b);
	if(n==1){
		head = head.next;
	}
	if(n>1){
		cur = head;
		while(--n != 1){
			cur = cur.next;
		}
		cur.next = cur.next.next;
	}
	return head;
}