Given a linked list, determine if it has a cycle in it.

Follow up:
Can you solve it without using extra space?

這道題是快慢指標的經典應用。只需要設兩個指標，一個每次走一步的慢指標和一個每次走兩步的快指標，如果連結串列裡有環的話，兩個指標最終肯定會相遇。實在是太巧妙了，要是我肯定想不出來。程式碼如下：

C++ 解法：

class Solution {
public:
    bool hasCycle(ListNode *head) {
        ListNode 
 
*slow = head, *fast = head;
        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
            if (slow == fast) return true;
        }
        return false;
    }
};

Java 解法：

public class Solution {
    public boolean
 
 hasCycle(ListNode head) {
        ListNode slow = head, fast = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            if (slow == fast) return true;
        }
        return false;
    }
}

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