Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
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. Note: 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.
Follow-up: Can you solve it without using extra space?
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
/*
suppose you have travel N nodes with slow then 2N with fast pointer. D is the distance between head and start of loop/cycle which we have to find and K is the distance from start of loop to node where slow and fast meet and Ci is the distance traveled in loop in i iteration by slow and cj for fast pointer.
N=D+k+ci
2N=D+k+cj
putting N value in 2 equ
2(D+k+ci)=D+k+cj
2D+2k+2ci=D+k+cj
D+k=c(j-2i)C
k=c(j-2i)C-D
So if we start from head to K(where they meet) we have to travel D distance and if we are D nodes behind K so we move D nodes from point where we met to reach the starting of loop
Follow Gaurav sen video for more explanation.
*/
public class Solution {
public ListNode detectCycle(ListNode head) {
ListNode slow=head;
ListNode fast=head;
if(head==null || head.next==null || head.next.next==null) return null;
slow=slow.next;
fast=fast.next.next;
while(slow!=fast && slow!=null && fast!=null && fast.next!=null){
slow=slow.next;
fast=fast.next.next;
}
if(slow==null || fast==null || fast.next==null) return null;
fast=head;
while(slow!=fast){
slow=slow.next;
fast=fast.next;
}
return slow;
}
}