判斷鏈表是否有環（Java實現）

判斷給定的鏈表中是否有環。如果有環則返回true，否則返回false。

解題思路：設置兩個指針，slow和fast，fast每次走兩步，slow每次走一步，如果有環的話fast一定會追上slow，判斷fast==slow或者fast.next==slow即可判斷

 class ListNode {
       int val;
       ListNode next;
       ListNode(int x) {
            val = x;
            next = null;
     }
 }


public class test1 {
    public boolean hasCycle(ListNode head) {
        if(head==null || head.next==null){
            //頭指針為空或者只有頭節點，無環
            return false;
        }
        ListNode slow,fast = new ListNode(0);
        slow = head.next;
        fast = head.next.next;
        while(true){
            if(fast==null||fast.next==null){
                //fast走到鏈表尾
                return false;
            }else if(fast.next==slow || fast==slow){
                return true;
            }else{
                slow = slow.next;// slow每次走一步
                fast = fast.next.next;//fast每次走兩步
            }
        }
    }

    public static void main(String[] args) {
        ListNode node1 = new ListNode(1),node2 = new ListNode(2),node3 = new ListNode(3),node4=new ListNode(4);
        node1.next=node2;
        node2.next=node3;
        node3.next=node4;
        node4.next=node1;
        test1 test = new test1();
        System.out.println(test.hasCycle(node1));
    }
}