题解 | #判断链表中是否有环#

//此题快慢指针可以很容易解决，只要有环，快慢指定必定会相遇
/**
 * 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) {
        return false;
    }
    ListNode fast = head;
    ListNode slow = head;
    while (fast != null && fast.next != null) {
        fast = fast.next.next;
        slow = slow.next;
        if (fast == slow) {
            return true;
        }

    }
    return false;

}

}