Question

142. Linked List Cycle II

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Description

Given the head of a linked list, return _the node where the cycle begins. If there is no cycle, return _null.

There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail's next pointer is connected to (0-indexed). It is -1 if there is no cycle. Note that pos is not passed as a parameter.

Do not modify the linked list.

 

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
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: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.

 

Constraints:

• The number of the nodes in the list is in the range [0, 104].
• -105 <= Node.val <= 105
• pos is -1 or a valid index in the linked-list.

 

Follow up: Can you solve it using O(1) (i.e. constant) memory?

Solutions

Solution 1: Two Pointers

We first use the fast and slow pointers to judge whether the linked list has a ring. If there is a ring, the fast and slow pointers will definitely meet, and the meeting node must be in the ring.

If there is no ring, the fast pointer will reach the tail of the linked list first, and return null directly.

If there is a ring, we then define an answer pointer $ans$ to point to the head of the linked list, and then let $ans$ and the slow pointer move forward together, moving one step at a time, until $ans$ and the slow pointer meet, and the meeting node is the ring entrance node.

Why can this find the entrance node of the ring?

Let's assume that the distance from the head node of the linked list to the entrance of the ring is $x$, the distance from the entrance of the ring to the meeting node is $y$, and the distance from the meeting node to the entrance of the ring is $z$. Then the distance traveled by the slow pointer is $x + y$, and the distance traveled by the fast pointer is $x + y + k \times (y + z)$, where $k$ is the number of times the fast pointer goes around the ring.

Because the speed of the fast pointer is twice that of the slow pointer, it is $2 \times (x + y) = x + y + k \times (y + z)$, which can be deduced that $x + y = k \times (y + z)$, that is $x = (k - 1) \times (y + z) + z$.

That is to say, if we define an answer pointer $ans$ to point to the head of the linked list, and the $ans$ and the slow pointer move forward together, they will definitely meet at the ring entrance.

The time complexity is $O(n)$, where $n$ is the number of nodes in the linked list. The space complexity is $O(1)$.

# Definition for singly-linked list.
# class ListNode:
#   def __init__(self, x):
#     self.val = x
#     self.next = None


class Solution:
  def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
    fast = slow = head
    while fast and fast.next:
      slow = slow.next
      fast = fast.next.next
      if slow == fast:
        ans = head
        while ans != slow:
          ans = ans.next
          slow = slow.next
        return ans

/**
 * Definition for singly-linked list.
 * class ListNode {
 *   int val;
 *   ListNode next;
 *   ListNode(int x) {
 *     val = x;
 *     next = null;
 *   }
 * }
 */
public class Solution {
  public ListNode detectCycle(ListNode head) {
    ListNode fast = head, slow = head;
    while (fast != null && fast.next != null) {
      slow = slow.next;
      fast = fast.next.next;
      if (slow == fast) {
        ListNode ans = head;
        while (ans != slow) {
          ans = ans.next;
          slow = slow.next;
        }
        return ans;
      }
    }
    return null;
  }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *   int val;
 *   ListNode *next;
 *   ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
  ListNode* detectCycle(ListNode* head) {
    ListNode* fast = head;
    ListNode* slow = head;
    while (fast && fast->next) {
      slow = slow->next;
      fast = fast->next->next;
      if (slow == fast) {
        ListNode* ans = head;
        while (ans != slow) {
          ans = ans->next;
          slow = slow->next;
        }
        return ans;
      }
    }
    return nullptr;
  }
};

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *   Val int
 *   Next *ListNode
 * }
 */
func detectCycle(head *ListNode) *ListNode {
  fast, slow := head, head
  for fast != nil && fast.Next != nil {
    slow = slow.Next
    fast = fast.Next.Next
    if slow == fast {
      ans := head
      for ans != slow {
        ans = ans.Next
        slow = slow.Next
      }
      return ans
    }
  }
  return nil
}

/**
 * Definition for singly-linked list.
 * class ListNode {
 *   val: number
 *   next: ListNode | null
 *   constructor(val?: number, next?: ListNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 *   }
 * }
 */

function detectCycle(head: ListNode | null): ListNode | null {
  let [slow, fast] = [head, head];
  while (fast && fast.next) {
    slow = slow.next;
    fast = fast.next.next;
    if (slow === fast) {
      let ans = head;
      while (ans !== slow) {
        ans = ans.next;
        slow = slow.next;
      }
      return ans;
    }
  }
  return null;
}

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *   this.val = val;
 *   this.next = null;
 * }
 */

/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var detectCycle = function (head) {
  let [slow, fast] = [head, head];
  while (fast && fast.next) {
    slow = slow.next;
    fast = fast.next.next;
    if (slow === fast) {
      let ans = head;
      while (ans !== slow) {
        ans = ans.next;
        slow = slow.next;
      }
      return ans;
    }
  }
  return null;
};