LeetCode 141. Linked List Cycle

141. Linked List Cycle

Given a linked list, determine if it has a cycle in it.

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.

Follow up:

Can you solve it using O(1) (i.e. constant) memory?

#include <iostream>

using namespace std;


struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};

class Solution
{
public:
    bool hasCycle(ListNode *head)
    {
        if (head == NULL) return false;
        ListNode *slow = head;
        ListNode *fast = head;
        while (fast && fast->next)
        {
            slow = slow->next;
            fast = fast->next->next;
            if (slow == fast) {
                return true;
            }
        }
        return false;
    }
};


void test_data()
{
    ListNode *head = new ListNode(0);
    ListNode *p;

    p = head;
    Solution s;

    int n = 0;
    int T = 5;

    while (T-- && cin >> n)
    {
        ListNode *q;
        q = new ListNode(n);
        p->next = q;
        p = q;
    }

    head = head->next;
    p->next = head;          //设置有环
    if (s.hasCycle(head)) {
        cout << "有环
";
    }
    else {
        cout << "无环
";
    }

}


void test_data1()
{
    ListNode *head = new ListNode(0);
    ListNode *p;

    p = head;
    Solution s;

    int n = 0;
    int T = 5;

    while (T-- && cin >> n)
    {
        ListNode *q;
        q = new ListNode(n);
        p->next = q;
        p = q;
    }

    head = head->next;        //不设置有环
    if (s.hasCycle(head)) {
        cout << "有环
";
    }
    else {
        cout << "无环
";
    }
}


int main()
{

    test_data();

    test_data1();

    return 0;

}