- 1、题目描述:
- 2、题目链接:
https://www.nowcoder.com/practice/650474f313294468a4ded3ce0f7898b9?tpId=117&&tqId=34925&rp=1&ru=/activity/oj&qru=/ta/job-code-high/question-ranking
-3、 设计思想:
详细操作流程看下图:
-4、视频讲解链接B站视频讲解
-5、代码:
c++版本:
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: bool hasCycle(ListNode *head) { if(head == NULL || head->next == NULL) return false; ListNode* fast = head->next;//快指针 ListNode *slow = head;//慢指针 while(slow != fast){//快慢指针不相遇就要遍历 if(fast == NULL || fast->next == NULL) return false; fast = fast->next->next;//快指针移动两格 slow = slow->next;//慢指针移动一格 } return true; } };
Java版本:
/** * 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 || head.next == null) return false; ListNode fast = head.next;//快指针 ListNode slow = head;//慢指针 while(fast != slow){//快慢指针不相遇就要遍历 if(fast == null || fast.next == null){ return false; } fast = fast.next.next;//快指针移动两格 slow = slow.next;//慢指针移动一格 } return true; } }
Python版本:
# class ListNode: # def __init__(self, x): # self.val = x # self.next = None # # # @param head ListNode类 # @return bool布尔型 # class Solution: def hasCycle(self , head ): # write code here if head == None or head.next == None: return False fast = head.next#快指针 slow = head#慢指针 while fast != slow:#快慢指针不相遇就要遍历 if fast == None or fast.next == None: return False fast = fast.next.next#快指针移动两格 slow = slow.next#慢指针移动一格 return True