Linkedlist_Cyle2.java

/*
	Linked List Cycle II 

	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?

*/

// Maintain two pointers to traverse the list
// Each time, slow pointer moves one step, fast pointer moves two steps
// remember key points:
//      <1> assume slow is k steps away from the head when it arrives entrance of the loop
//          ,then fast is k steps ahead of the slow (or L-k steps behind the slow, 
//          L is the length of the cycle).
//      <2> fast can catch up with slow in L-k steps
//      <3> when fast catches slow, slow is L-k steps ahead of the entrance (so as fast).
//          This means that it will take fast k steps to get to the entrance again.
// Another explaination:
//      <1> when slow meets fast, fast is k steps away from slow, which is also the length of the cycle
//      <2> L = A+B, K = C + B, if L==K, A==C
// time: O(n); space: O(1)

/**
 * 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) {
        if(head==null) return null;
        ListNode slow=head;
        ListNode fast=head;
        while(fast!=null && slow!=null && fast.next!=null){
            slow=slow.next;
            //fast=fast.next.next;
            if(fast.next.next==null) return null;
            else fast=fast.next.next;
            if(fast==slow)  break;  //This isn't the begin of cycle.
        }
        if(slow==null || fast==null || fast.next==null) return null; // 检查是否因为有环退出或是因为碰到null而退出
        fast=head;
        while(fast!=slow){  // 再次相遇时即为环的入口
            fast=fast.next;
            slow=slow.next;
        }
        return fast;
    }
}