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/**
 * Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
 */
 
public class Solution{
    public ListNode detectCycle(ListNode head) {
        // The slow pointer advances one node at a time, while the fast pointer
    	// traverses twice as fast. If the list has loop in it, eventually the
		// fast and slow pointer will meet at the same node. On the other hand,
		// if the loop has no loop, the fast pointer will reach the end of list
		// before the slow pointer does.
        if(head == null || head.next == null || head.next.next == null)
            return null;
        ListNode slow = head.next;
        ListNode fast = head.next.next;
        while(slow != fast){
            if(fast.next == null || fast.next.next == null)
                return null;
            slow = slow.next;
            fast = fast.next.next;
        }
        // If the program runs here, which means there is a loop, now advance
    	// one pointer to the head, and make them move in the same speed.
		// When they meet again, it has to be the loop start point.
        slow = head; // or fast = head
        while(slow != fast){
            slow = slow.next;
            fast = fast.next;
        }
        return slow;
    }
}

/**
 * Given a linked list, determine if it has a cycle in it.
 */
 
/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution{
    public boolean hasLoop(ListNode head){
        ListNode slow = head, fast = head;
        while(slow != null && fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
            if(slow == fast)
                return true;
        }
        return false;
    }
}