# 142. Linked List Cycle II
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?
# Solution
Approach 1: Hash table.
Approach 2: Two pointers.
# Code (Python)
Approach 1:
Approach 2:
# Code (C++)
Approach 1:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
// Hash table.
class Solution {
public:
ListNode *detectCycle(ListNode *head) {
unordered_set<ListNode*> visited;
ListNode *curr = head;
while (curr)
{
if (visited.find(curr) != visited.end())
return curr;
else
visited.insert(curr);
curr = curr->next;
}
return NULL;
}
};
Approach 2:
// Two pointers. When fast and slow pointers meet, slow pointer
// has moved x steps while fast pointer has moved 2x steps.
// Now let slow pointer starts from the list head and fast pointer
// keeps moving with one step each time. They will meet in the same
// node again after x steps. Meanwhile, the first node they meet is
// where the cycle begins.
class Solution {
public:
ListNode *detectCycle(ListNode *head) {
ListNode *slow = head;
ListNode *fast = head;
while (fast && fast->next)
{
slow = slow->next;
fast = fast->next->next;
if (slow == fast)
break;
}
slow = head;
while (fast && fast->next)
{
if (slow == fast)
return slow;
slow = slow->next;
fast = fast->next;
}
return NULL;
}
};