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 ListNode
class 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