Given a linked list, return the node where the cycle begins. If there is no cycle, return null
.
Follow up:
Can you solve it without using extra space?
Analysis:
Definitions:
Cycle = length of the cycle, if exists.
C is the beginning of Cycle, S is the distance of slow pointer from C when slow pointer meets fast pointer.
Distance(slow) = C + S, Distance(fast) = 2 * Distance(slow) = 2 * (C +
S). To let slow poiner meets fast pointer, only if fast pointer run 1
cycle more than slow pointer. Distance(fast) - Distance(slow) = Cycle
=> 2 * (C + S) - (C + S) = Cycle
=> C + S = Cycle
=> C = Cycle - S
=> This means if slow pointer runs (Cycle - S) more, it will reaches
C. So at this time, if there's another point2 running from head
=> After C distance, point2 will meet slow pointer at C, where is the beginning of the cycle.
Solution:
/** * 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 slow = head, fast = head; while (fast != null && fast.next!=null){ fast = fast.next.next; slow = slow.next; if (slow==fast){ ListNode p2 = head; while (p2!=slow){ slow = slow.next; p2 = p2.next; } return slow; } } return null; } }