141. Linked List Cycle

Problem statement

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

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

Solution

This is a classical problem in a linked list. It is chosen for interview by many high-tech companies.

The generally idea is also a very good philosophy in linked list -->  Fast and slow pointers.

• Define two pointers, fast and slow.
• Each time fast goes two step and slow goes one step.
• If there is a loop, they must met sometime.
• Otherwise, fast will become NULL first.

Time complexity is O(n), space complexity is O(1)

The following solution follows the idea to find the middle of a link list. Fast pointer should goes faster than slow at start.

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

This solution do not need to do the test at beginning of the program, however, fast is initialized the same value of slow.

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