題目

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

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

思路

本來我傻傻的用  Linked List Cycle 的方法來做，以為找到了節點就意味著這個點就是環的開始。但是，結果報了個錯

於是換了另一個思路.

我們假設slow走了n步，而fast走了2n步.在環的某個點相遇（假設點D).那麼，想一下.如果讓fast不動,讓slow在走n步,slow是不是剛好又回到了這個相遇點（D）？（因為fast走了2n步到達這個地方).那麼此時，如果讓fast也從連結串列的開頭走，走n步.那麼fast和slow必定會在D相遇.但是，這之前，它們就會提前相遇，而相遇點就是環入口！

程式碼

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode detectCycle(ListNode head) {
		ListNode node = null;
		if(head == null)
			return node;
		ListNode quick = head;
		ListNode slow = head;
		
		boolean tag = false;
		while(quick!=null &&quick.next != null)
		{
			quick = quick.next;
			quick = quick.next;
			slow = slow.next;
			if(quick == slow)
			{
				tag = true;
				break;
			}
		} 
		
		if( tag  == false)
		{
			return node;
		}
		else
		{
			quick = head;
			while( true)
			{
				if( quick == slow)
				{
					return quick;
				}
				else
				{
					quick = quick.next;
					slow = slow.next;
				}
			}
		}
    }
}