It returns the list of intermediate nodes, if there are two intermediate nodes, the latter returns a point (two ways) junction

1. First traverse the entire list, obtains the length of the list, identify an intermediate node during the cycle

public class Solution {
    public ListNode middleNode(ListNode head) {
		int len=0;
		for(ListNode cur=head;cur!=null;cur=cur.next)
			len++;//遍历算出链表长度
		ListNode node=head;
		for(int i=0;i<len/2;i++)
			node=node.next;
		return node;
	}
}

2. Use speed pointers define two references, FAST pointer moving twice, slow pointer a step, until the stop fast.next = null, i.e. an intermediate node

public class Solution {
    public ListNode middleNode(ListNode head) {
		ListNode fast=head;
		ListNode slow=head;
		while(fast!=null){
			fast=fast.next;
			if(fast==null){
				break;
			}//因为fast每次走两步，有可能走一步时就到了最后（奇数节点时）
			slow=slow.next;
			fast=fast.next;
		}
		return slow;
	}
}