142. Linked List Cycle II

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142. Linked List Cycle II

Total Accepted: 70643 Total Submissions: 224496 Difficulty: Medium

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?

现在有两个指针，第一个指针，每走一次走一步，第二个指针每走一次走两步，如果他们走了t次之后相遇在K点

那么       指针一  走的路是      t = X + nY + K        ①

             指针二  走的路是     2t = X + mY+ K       ②          m,n为未知数

把等式一代入到等式二中, 有

2X + 2nY + 2K = X + mY + K

=>   X+K  =  (m-2n)Y    ③

这就清晰了，X和K的关系是基于Y互补的。等于说，两个指针相遇以后，再往下走X步就回到Cycle的起点了。这就可以有O(n)的实现了。

/**
 * Linked List Cycle II
 * 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.
 * Tag : Linked List Two Pointers
 * Similar Problems: (M) Linked List Cycle (H) Find the Duplicate Number
 * 
 * Analysis:
 * fast point: 2d = x + y + mC
 * slow point: d = x + y + nC
 * x + y = (m - 2n) * C*/

public class LinkedListCycle_2 {
    public static ListNode detectCycle(ListNode head) {
        if (head == null || head.next == null) {
            return null;
        }  
        ListNode fast = head;
        ListNode slow = head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            if (fast == slow) {
                break;
            }
        }
        if (slow == fast) {
            fast = head;
            while (slow != fast) {
                slow = slow.next;
                fast = fast.next;
            }
            return slow;
        }
        else {
            return null;
        }   
    }
    
    public static void main(String[] args) {
        ListNode res = new ListNode(1);
        res.next = new ListNode(2);
        res.next.next = new ListNode(3);
        res.next.next.next = new ListNode(4);
        res.next.next.next.next = new ListNode(5);
        res.next.next.next.next.next = new ListNode(6);
        res.next.next.next.next.next = res.next.next.next;
        System.out.print(detectCycle(res).val);    
    }
}
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