"Leetcode from zero single brush" Linked list Cycle

Topic:

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

Can you solve it without using extra space?

Answer:

At first glance it's simple. My practice is to save the head pointer, and then the other pointer continues through until the head pointer is encountered. But I did not take into account the partial loopback situation:

To tell the truth, I didn't think of any good way. Until the discussion, found that there can be a pursuit of the problem of the idea to do: a slow pointer to run a step at a time, a quick pointer run two steps at a time. If there is a ring must be able to catch up.

So I wrote down:

Class Solution {public:    bool Hascycle (ListNode *head) {        if (head = = NULL | | head->next = NULL | | head->next- >next = = NULL)            return false;                listnode* slow = head->next;        listnode* fast = head->next->next;        while (slow! = NULL)        {            if (slow = = fast)            {                return true;            }                        if (Fast->next->next = = NULL)            {                return false;            }            else            {                slow = slow->next;                Fast = fast->next->next;            }        }        return false;    }};

Show Runtime Error ... Later found, when judging if (Fast->next->next = = null), I forgot to consider: if Fast->next = = NULL, then Fast->next->next became a wild pointer /c23>, and then else, if you assign a field pointer to fast, there's no end to it.

The IF (Fast->next->next = = null) is determined to be changed to if (Fast->next->next = = NULL | | fast->next = = NULL), or wrong ...

Then the basic skills exposed ... The condition "or" operation is a short-circuit operation . and Fast->next = = NULL is a condition that needs to be judged in advance , so it should be judged beforehand. Final AC version:

Class Solution {public:    bool Hascycle (ListNode *head) {        if (head = = NULL | | head->next = NULL | | head->next- >next = = NULL)            return false;                listnode* slow = head->next;        listnode* fast = head->next->next;        while (slow! = NULL)        {            if (slow = = fast)            {                return true;            }                        if (Fast->next = = NULL | | fast->next->next = = NULL)            {                return false;            }            else            {                slow = slow->next;                Fast = fast->next->next;            }        }        return false;    }};

"Leetcode from zero single brush" Linked list Cycle