LintCode（102） 带环链表

题目

给定一个链表，判断它是否有环。

您在真实的面试中是否遇到过这个题？ 

Yes

样例

给出 -21->10->4->5, tail connects to node index 1，返回 true

分析

判断链表有无环的问题，很经典，也是面试中常见的问题。

快慢指针解决。

Python代码

"""Definition of ListNodeclass ListNode(object):    def __init__(self, val, next=None):        self.val = val        self.next = next"""class Solution:    """    @param head: The first node of the linked list.    @return: True if it has a cycle, or false    """    def hasCycle(self, head):        # write your code here        if head == None or head.next == None:            return False                slow = head        fast = head        while fast != None and fast.next != None:            slow = slow.next            fast = fast.next.next            if slow == fast:                return True                        return False

GitHub -- Python代码

C++代码

/**102 带环链表给定一个链表，判断它是否有环。您在真实的面试中是否遇到过这个题？ Yes样例给出 -21->10->4->5, tail connects to node index 1，返回 true*//** * Definition of ListNode * class ListNode { * public: *     int val; *     ListNode *next; *     ListNode(int val) { *         this->val = val; *         this->next = NULL; *     } * } */class Solution {public:    /**     * @param head: The first node of linked list.     * @return: True if it has a cycle, or false     */    bool hasCycle(ListNode *head) {        // write your code here        if(head == NULL || head->next == NULL)        {            return false;        }//if                ListNode *slow = head, *fast = head;        while(fast != NULL && fast->next != NULL)        {            fast = fast->next->next;            slow = slow->next;            if(slow == fast)            {                return true;            }//if        }//while        return false;    }};

GitHub -- C++代码