Algorithm
Problem Name: 142. Linked List Cycle II
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 <= 105posis-1or a valid index in the linked-list.
Code Examples
#1 Code Example with C Programming
Code -
C Programming
struct ListNode *detectCycle(struct ListNode *head) {
struct ListNode *n, *nn;
if (!head) return NULL;
n = nn = head;
do {
n = n->next; // one step
nn = nn->next;
if (nn) nn = nn->next; // two steps
} while (nn && n != nn);
if (!nn) return NULL;
// there is a cycle
n = head;
while (n != nn) {
n = n->next;
nn = nn->next;
}
return n;
}Input
Output
#2 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
ListNode *detectCycle(ListNode *head) {
auto one = head, two = head, meet = head;
while(two && two->next){
one = one->next;
two = two->next->next;
if(one == two){
meet = one;
break;
}
}
if(!two || !two->next) return NULL;
auto p = head;
while(p != meet){
p = p->next;
meet = meet->next;
}
return p;
}
};
Input
Output
#3 Code Example with Java Programming
Code -
Java Programming
public class Solution {
public ListNode detectCycle(ListNode head) {
ListNode fast = head;
ListNode slow = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
if (slow == fast) {
break;
}
}
if (fast == null || fast.next == null) {
return null;
}
slow = head;
while (slow != fast) {
slow = slow.next;
fast = fast.next;
}
return slow;
}
}
Input
Output
#4 Code Example with Javascript Programming
Code -
Javascript Programming
const detectCycle = function(head) {
if(head === null || head.next === null) return null
let fast = head
let slow = head
let start = head
while(fast !== null && fast.next !== null) {
fast = fast.next.next
slow = slow.next
if(fast === slow) {
while(slow !== start) {
slow = slow.next
start = start.next
}
return start
}
}
return null
};
Input
Output
#5 Code Example with Python Programming
Code -
Python Programming
class Solution:
def detectCycle(self, head: ListNode) -> ListNode:
fast = slow = root = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
if slow == fast:
while root != slow:
root = root.next
slow = slow.next
return root
Input
Output