Find cycle in a linked list. (Method 1)

Question: Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

Output: 0

We are given a linked list which may or may not have a loop. A loop in a linked list means that if we keep doing a next, we will never reach null. Thus, we can be clear of a fact that if we reach null, then the list does not have a loop.

So how do we detect if there is a loop in the linked list. One method that immediately comes to our mind is that if a value is encountered twice, there may be a chance that the linked list has a loop. Thus, we can use Sets to keep track of all the values traversed.

Algorithm

• Allocate a Set to store ListNode references.
• Traverse the list, checking visited for containment of the current node.
• If the node has already been seen, then it is necessarily the entrance to the cycle.
• If any other node were the entrance to the cycle, then we would have already returned that node instead.
• Otherwise, the if condition will never be satisfied, and our function will return null.

The algorithm necessarily terminates for any list with a finite number of nodes, as the domain of input lists can be divided into two categories: cyclic and acyclic lists. An acyclic list resembles a null-terminated chain of nodes, while a cyclic list can be thought of as an acyclic list with the final null replaced by a reference to some previous node. If the while loop terminates, we return null, as we have traversed the entire list without encountering a duplicate reference. In this case, the list is acyclic. For a cyclic list, the while loop will never terminate, but at some point the if condition will be satisfied and cause the function to return.

// Template for a linked list node
class ListNode {
int value;
ListNode next;
}

public class Solution {

Set<ListNode> visited = new HashSet<ListNode>();

// Traverse the list
while (node != null) {

// If the node is already present in the Set
// this is a loop and we need to return it.
if (visited.contains(node)) {
return node;
}

node = node.next;
}
// No loop found
return null;
}
}
Code language: Java (java)

Time Complexity: O(n)
Space Complexity: O(n)

A sample problem can be found here.

A method with constant space can be found here.

1 comment

June 5, 2020 - 03:39

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