leetcode刷题系列----模式1（Two Points 双指针）- 141：Linked List Cycle环形链表

• Java和C#核心代码完全一样。
• 这题比较简单，类似于时针和分针何时相遇。
• 注意算法与数据结构的定义高度绑定，一定参照如何定义链表来思考解决此类问题的算法。

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def hasCycle(self, head: ListNode) -> bool:
        if not head or not head.next:
            return False
        
        fast = low = head
        while fast:
            fast = fast.next
            if fast:
                fast=fast.next
            low = low.next
            if fast == low:
                return True
        return False

class Solution {
public:
    bool hasCycle(ListNode *head) {
        if(head==nullptr || head->next==nullptr) return false;
        
        ListNode* fast = head;
        ListNode* low = head;
        while(fast!=nullptr)
        {
            fast = fast->next;
            if(fast==low) return true;
            
            if(fast!=nullptr) fast=fast->next;
            low = low->next;
            
        }
        return false;
    }
};

public class Solution {
    public bool HasCycle(ListNode head) {
        if(head==null || head.next==null) return false;
        ListNode slow = head;
        ListNode fast = head;
        while(fast!=null)
        {
            fast = fast.next;
            if(fast!=null) fast = fast.next;
            if(slow==fast) return true;
            slow = slow.next;
        }
        return false;
    }
}

public class Solution {
    public boolean hasCycle(ListNode head) {
        if(head==null || head.next==null) return false;
        ListNode slow = head;
        ListNode fast = head;
        while(fast!=null)
        {
            fast = fast.next;
            if(fast!=null) fast = fast.next;
            if(slow==fast) return true;
            slow = slow.next;
        }
        return false;
    }
}