[Leetcode]142. Linked List Cycle II

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

Note: Do not modify the linked list.

思路：我的想法是先用快慢指针判断是否有环，如果有环的话，找出快慢指针相遇的那个点。

然后让一个指针一直在那个环里循环，而另一个新指针从头节点开始走，返回他们相遇的那个点；

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    private ListNode cur;
    public ListNode detectCycle(ListNode head) {
        if (head==null||head.next==null)
            return null;
        ListNode p = head,q = head;
        while (q!=null){
            q = q.next;
            if (q==null)
                return null;
            q = q.next;
            p = p.next;
            if (p==q)
                break;
        }
        if (q==null)
            return null;
        else 
            helper(head,p,q.next);
        return cur;
    }
    private void helper(ListNode head,ListNode p,ListNode q){
        if (p==head){
            cur = head;
        }
        if (cur==null){
            while(q!=p){    
                p = p.next; 
                if (p==head)
                    cur = head;
            }
            helper(head.next,p,q.next);
        }
    }
}