面试题23：链表环中的入口节点

public class Solution {

    public ListNode EntryNodeOfLoop(ListNode pHead) {
        ListNode meetNode = meetNode(pHead);
        if (meetNode == null) {
            return null;
        }
        //计算环的节点数，用相遇节点作为起点，从前往后走，计数即可
        //cicleCount从1开始，因为有只有一个节点的环{}{1}，即next指向自己，对应的输出应该为1
        int cicleCount = 1;
        ListNode pNode = meetNode;
        while (pNode.next != meetNode) {
            cicleCount++;
            pNode = pNode.next;
        }
        //定义两个指针，第一个指针原位，第二个指针提前向前走cicleCount步
        ListNode aheadNode = pHead;
        ListNode behindNode = pHead;
        for (int i = 0; i < cicleCount; i++) {
            aheadNode = aheadNode.next;
        }
        //求环入口节点，两个指针一起走，相遇的就是入口节点
        while (aheadNode != behindNode) {
            aheadNode = aheadNode.next;
            behindNode = behindNode.next;
        }
        return behindNode;
    }

    public static ListNode meetNode(ListNode pHead) {
        if (pHead == null) {
            return null;
        }

        ListNode slowNode = pHead.next;
        if (slowNode == null) {
            return null;
        }
        ListNode fastNode = slowNode.next;

        while (slowNode != null && fastNode != null) {
            if (slowNode == fastNode) {
                return slowNode;
            } else {
                slowNode = slowNode.next;
                if (slowNode == null) {
                    break;
                }
                fastNode = fastNode.next;
                if (fastNode != null) {
                    fastNode = fastNode.next;
                }

            }
        }
        return null;
    }
}