链表中是否有环以及环的位置

package swordoffer;

public class Loop <AnyType>{
	private static class Node <AnyType>{
		public Node<AnyType> next ;
		public AnyType data ;
		
		public Node(Node<AnyType> next , AnyType data){
			this.next = next ;
			this.data = data ;
		}
	}
	boolean hasCircle (Node<AnyType> head ){
		Node fast = head ,slow = head ;
		Node<AnyType> encounter ;
		
		while (fast!=null && fast.next!= null){
			fast = fast.next.next ;
			slow = slow.next ;
			if (fast== slow){
				encounter = fast ;//用在后面
				return true ;
			}
		}
		encounter = null ;
		return false ;
	}
	
	/**
	 * s +nr =2s 
	 * s = x+y
	 * 得到
	 * nr = x+y 
	 * x =nr-y 
	 * 也就是说，p1 从链表 开始处遍历，p2从encounter处遍历，一次都 移动 一步，则当p1到达入口点时，p2移动nr-y步
	 * 两者恰好在环的入口 点处相遇了。
	 * @param head
	 * @param encounter
	 * @return
	 */
	Node<AnyType> findEntry (Node<AnyType> head , Node<AnyType> encounter) {
		Node<AnyType> p1 = head , p2 = encounter ;
		while(p1!= p2 ){
			p1= p1.next ;
			p2 = p1.next ;
		}
		return p1 ;
	}

}

参考 了

http://blog.csdn.net/wuzhekai1985/article/details/6725263