Sitrone
diff --git a/‎solutions/141. Linked List Cycle/AC_Floyd I.java‎
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diff --git a/‎solutions/141. Linked List Cycle/AC_Floyd II.java‎
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diff --git a/‎solutions/141. Linked List Cycle/notes.md‎
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+	/**
+	 * 1、Floyd判圈法（之前一直叫双指针跑圈法来着）
+	 * 
+	 *
+	 */
+/**
+ * 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;
+		do{
+			if(fast == null || fast.next == null) return false;
+			
+			slow = slow.next;
+			fast = fast.next.next;
+		}while(slow != fast);
+		return true;
+    }
+}
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+	/**
+	 * 在Linded List Cycle I的基础上再求出环的起始节点
+	 * 
+	 *
+	 */
+/**
+ * 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) {
+        if(head == null || head.next == null)
+			return null;
+		
+		ListNode slow = head;
+		ListNode fast = head;
+		do{
+			if(fast == null || fast.next == null) return null;
+			
+			slow = slow.next;
+			fast = fast.next.next;
+		}while(slow != fast);
+
+		ListNode start = head;
+		while(slow != start){
+			slow = slow.next;
+			start = start.next;
+		}
+		return start;
+    }
+}
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+###参考
+1. [wiki Floyd](https://en.wikipedia.org/wiki/Cycle_detection#Tortoise_and_hare)
+
+###判断链表有环的相关问题：
+1、链表是否有环
+> 双指针，一个slow一次一步，一个fast一次两步，如果相遇，则有环；如果fast指向null，则无环。
+
+[Leetcode 问题:141. Linked List Cycle](https://leetcode.com/problems/linked-list-cycle/)
+
+2、如果有环，求环的长度
+> 接上，双指针，第一次相遇后开始计数，再次相遇，则为环长
+
+3、求环的起点
+> 设初始起点到环的起点距离为**m**，环的长度为**n**，第一次相遇时候距离环的起点为**k**
+  第一次相遇slow跑的节点数 i = m + a * n + k   （1）
+  第一次相遇fast跑的节点数 2i = m + b * n + k   （2）
+  （2） - （1）： i = (b - a) * n
+  可知慢节点跑的距离为环长的整数倍，即 m + k = c * n,也就是：
+  如重新设置一个新的指针指向起始节点new，开始计数，当new与slow相遇时候，这个点就是环的起点
+  
+[Leetcode 问题:142: Linked List Cycle II](https://leetcode.com/problems/linked-list-cycle-ii/)