Difficulty:medium

 More:【目录】LeetCode Java实现

Description

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

Note: Do not modify the linked list.

Follow up:
Can you solve it without using extra space?

Intuition

1) find out whether there is a cycle, refer to141. Linked List Cycle

2) if there is a cycle, we can get the meetingNode, meanwhile we use a pointer named entry to point to the head Node.

3) let the meetingNode and entry move one step at a time, where they meet is where the cycle begins.

why will they meet at the beginning of the cycle? here is the proof:

  1. define L1 as the distance between the head node and entry node;

  2.defined L2 as the distance between the entry node and meeting node;

  3.defined C as the length of the cycle

  We can know that:

   the distance of slow node=L1+L2; while the distance of fast node=L1+L2+n*C

  because of the speed, we have :

  2*(L1+L2)=L1+L2+n*C => L1=n*C-L2 => L1=(n-1)C+(C-L2)

  it means that: the distance between the head node and entry node equals the distance between the meeting node and entry node;

Solution

    public ListNode detectCycle(ListNode head) {
if(head==null)
return null;
ListNode fast=head;
ListNode slow=head;
ListNode entry=head;
while(fast.next!=null && fast.next.next!=null){
fast=fast.next.next;
slow=slow.next;
if(fast==slow){
while(entry!=slow){
entry=entry.next;
slow=slow.next;
}
return entry;
}
}
return null;
}

  

Complexity

Time complexity : O(n)

Space complexity : O(1)

What I've learned

1. How to prove the distance relation.

 More:【目录】LeetCode Java实现

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