142. Linked List Cycle II

fast ans slow pointer Solution

• We can use fast and slow pointer method to find the cycle.
• We move fast by 2 steps and slow by 1 step.
• When they both are equal, we have found the cycle, else we return null.
• If cycle found set fast pointer to head again and move both by 1 step.
• when both of then are equal, we have found the start of the cycle.
• return the fast/slow pointer.
• TC: O(N)
• SC: O(1)

class Solution {
public:
    ListNode* detectCycle(ListNode* head)
    {
        // base case
        if (head == NULL || head->next == NULL) return NULL;

        ListNode *slow = head, *fast = head;
        while (fast->next != NULL && fast->next->next != NULL) {
            slow = slow->next;
            fast = fast->next->next;
            if (slow == fast) {
                fast = head;
                while (slow != fast) {
                    slow = slow->next;
                    fast = fast->next;
                }
                return fast;
            }
        }
        return NULL;
    }
};