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

Follow up:
Can you solve it without using extra space?

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
bool hasCycle(struct ListNode *head) 
{
    if(NULL == head)
        return false;
    
    struct ListNode *slow = head;
    struct ListNode *fast = head->next;
    
    while(slow && fast)
    {
        slow = slow->next;
        fast = fast->next;
        
        if(fast!=NULL)
            fast = fast->next;
        
        if(slow == fast)
            return true;
    }
    
    return false;
    
}

更为清晰的解法：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
bool hasCycle(struct ListNode *head) 
{
    if(NULL == head)
        return false;
    
    struct ListNode *slow = head;
    struct ListNode *fast = head;
    
    while(slow && fast && fast->next)
    {
        slow = slow->next;
        fast = fast->next->next;
        
        if(slow == fast)
            return true;
    }
    
    return false;
    
}

C++解法，快慢指针初始化为相同起点：

/**
 * 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) {
        ListNode *slow = head;
        ListNode *fast = head;
        while(fast && fast->next){
            slow = slow->next;
            fast = fast->next->next;
            if(slow == fast) return true;
        }
        
        return false;
    }
};

C++解法，快慢指针初始化为不同起点：

/**
 * 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 == nullptr || head->next == nullptr)
            return false;
        ListNode *slow = head;
        ListNode *fast = head->next;
        while(slow&&fast){
            if(slow == fast) return true;
            slow = slow->next;
            fast = fast->next;
            if(fast) fast = fast->next;
            else return false;
        }
        
        return false;
    }
};