LeetCode -- Linked List Circle ii

Question:

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

 

Analysis:

只想到了，首先判断是否有环，若有环，则总链首开始，一次判断是否是环的开始，这样T(O) = O(n^2)。

其实是一个数学问题，详细思路参照链接http://blog.csdn.net/sbitswc/article/details/27584037 。

 

Answer：

    
    public ListNode detectCycle(ListNode head) {
        ListNode fast = head, slow = head;
        while(fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            if(fast == slow)
                break;
        }
        
        if(fast == null || fast.next == null)
            return null;
        
        slow = head;
        while(fast != slow) {
            fast = fast.next;
            slow = slow.next;
        }
        return fast;
    }