142. Linked List Cycle II (Medium)
https://leetcode.com/problems/linked-list-cycle-ii/
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?
Solutions
public class Solution {
// |----x1-----|--p-x2---|
// x1 represent the normal section of the list without circle, x2 represents the circle section of the list
// Assume step length of slow is 1, fast encounters slow after slow walks S steps.
// We can derive that the length slow walked is S, and fast is 2*S
// Suppose fast and slow meets at point p, subject to (2*S-x1)%x2 = (S-x1)%x2=p-x1
// A simplified formula of above is S%x2=0
// Then slow continues to proceed forward x1 steps, the finally position is x1+(S+x1-x1)%x2=?
// The simplified equation is x1+S%x2==x1+0=x1, that explains why after walking x1 steps(x1 length)
// slow2 meets slow at the start of the circle
public ListNode detectCycle(ListNode head) {
ListNode slow = head;
ListNode fast = head;
while (fast != null && fast.next != null) {
slow = slow.next;
// even step size
fast = fast.next.next;
if (fast != slow) {
continue;
}
// fast == slow means both pointer fast and slow already run into a circle
// and the pointer fast and slow can be any position on the circle
ListNode slow2 = head;
while (slow2 != slow) {
slow = slow.next;
slow2 = slow2.next;
}
return slow;
}
return null;
}
}