My code:

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public boolean hasCycle(ListNode head) {
        if (head == null)
            return false;
        ListNode tortoise = head;
        ListNode rabbit = head.next;
        while (rabbit != null) {
            if (rabbit == tortoise)
                return true;
            rabbit = rabbit.next;
            if (rabbit == null)
                return false;
            rabbit = rabbit.next;
            tortoise = tortoise.next;
        }
        return false;
    }
}

My test result:

Paste_Image.png

这道题目不难，就是龟兔赛跑，快慢指针。
一个指针排在前面，每次走两步，一格指针摆在后面一格，每次走一步。
如果有环，进入环后，快的指针一定可以赶上慢的，或者说，碰到。

**
总结： 快慢指针， LinkedList
**

Anyway, Good luck, Richardo!

My code:

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public boolean hasCycle(ListNode head) {
        if (head == null || head.next == null)
            return false;
        ListNode slow = head;
        ListNode fast = head;
        while (fast.next != null && fast.next.next != null) {
            slow = slow.next;
            fast = fast.next;
            fast = fast.next;
            if (slow == fast)
                return true;
        }
        return false;
    }
}

快慢指针。

Anyway, Good luck, Richardo!

差不多的做法。
Anyway, Good luck, Richardo! -- 08/15/2016