题目描述：
在本题中，单链表可能有环，也可能无环。给定两个单链表的头节点 head1和head2，这两个链表可能相交，也可能不相交。请实现一个函数， 如果两个链表相交，请返回相交的第一个节点；如果不相交，返回null 即可。 要求：如果链表1的长度为N，链表2的长度为M，时间复杂度请达到 O(N+M)，额外空间复杂度请达到O(1)。

public class Code_022_FindFirstIntersectNode {
	
	public static class Node {
		public int data;
		public Node next;
		
		public Node(int data) {
			this.data = data;
		}
	}
	
	public static Node getIntersectNode(Node head1, Node head2) {
		if (head1 == null || head2 == null) {
			return null;
		}
		
		Node loop1 = getLoopNode(head1);
		Node loop2 = getLoopNode(head2);
		//两个链表都是无环链表的相交情况
		if (loop1 == null && loop2 == null) {
			return noLoop(head1, head2);
		}
		//两个链表都是有环链表的相交情况
		if (loop1 != null && loop2 != null) {
			return bothLoop(head1, loop1, head2, loop2);
		}
		
		//一个链表有环,一个链表无环,不可能相交
		return null;
	}
	
	//判断一个链表是否有环,如果有返回入环的第一个节点,否则返回null
	public static Node getLoopNode(Node head) {
		if (head == null) {
			return null;
		}
		
		Node n1 = head.next;	//慢指针
		Node n2 = head.next.next;	//快指针
		while(n1 != n2) {
			if (n2.next == null || n2.next.next == null) {
				return null;
			}
			n1 = n1.next;
			n2 = n2.next.next;
		}
		n2 = head;
		while(n1 != n2) {
			n1 = n1.next;
			n2 = n2.next;
		}
		
		return n1;
	}
	
	//判断两个无环链表是否相交,如果相交返回相交的第一个节点,否则返回null
	public static Node noLoop(Node head1, Node head2) {
		if (head1 == null || head2 == null) {
			return null;
		}
		
		int n = 0;
		Node cur1 = head1;
		Node cur2 = head2;
		while(cur1.next != null) {
			n++;
			cur1 = cur1.next;
		}
		while(cur2.next != null) {
			n--;
			cur2 = cur2.next;
		}
		if (cur1 != cur2) {
			return null;
		}
		cur1 = n > 0 ? head1 : head2;
		cur2 = cur1 == head1 ? head2 : head1;
		n = Math.abs(n);
		while(n > 0) {
			n--;
			cur1 = cur1.next;
		}
		while(cur1 != cur2) {
			cur1 = cur1.next;
			cur2 = cur2.next;
		}
		
		return cur1;
	}
	
	//判断两个有环链表是否相交,如果相交返回相交的第一个节点,否则返回null
	public static Node bothLoop(Node head1, Node loop1, Node head2, Node loop2) {
		Node cur1 = null;
		Node cur2 = null;
		if (loop1 == loop2) {
			int n = 0;
			cur1 = head1;
			cur2 = head2;
			while(cur1 != loop1) {
				n++;
				cur1 = cur1.next;
			}
			while(cur2 != loop2) {
				n--;
				cur2 = cur2.next;
			}
			if (cur1 != cur2) {
				return null;
			}
			cur1 = n > 0 ? head1 : head2;
			cur2 = cur1 == head1 ? head2 : head1;
			n = Math.abs(n);
			while(n != 0) {
				n--;
				cur1 = cur1.next;
			}
			while(cur1 != cur2) {
				cur1 = cur1.next;
				cur2 = cur2.next;
			}
			
			return cur1;
		} else {
			cur1 = loop1.next;
			while(cur1 != loop1) {
				if (cur1 == loop2) {
					return loop1;
				}
				cur1 = cur1.next;
			}
			
			return null;
		}
	}
}