NC4 判断链表中是否有环

python3：

class Solution:
    def hasCycle(self , head ):
        # write code here
        if not head:
            return head
        # 双指针  快慢指针
        slow = head
        fast = head
        while slow and fast:
            slow = slow.next
            if fast.next:
                fast = fast.next.next
            else:
                return False
            # 当双指针相遇 即表示指针有环
            if slow == fast:
                return True
        return False

java：

public class Solution {
    public boolean hasCycle(ListNode head) {
        if (head == null)
            return false;
        //快慢两个指针
        ListNode slow = head;
        ListNode fast = head;
        while (fast != null && fast.next != null) {
            //慢指针每次走一步
            slow = slow.next;
            //快指针每次走两步
            fast = fast.next.next;
            //如果相遇，说明有环，直接返回true
            if (slow == fast)
                return true;
        }
        //否则就是没环
        return false;
    }
}

快慢指针问题即可解决，当两个指针相遇，即证明链表有环