142 Linked List Cycle II

142. Linked List Cycle II

1. Question

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

Note:Do not modify the linked list.

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

2. Implementation

/**
 * 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 || head.next == null) {
            return null;
        }

        ListNode slow = head, fast = head;

        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;

            if (fast == slow) {
                break;
            }
        }

        if (fast != slow) {
            return null;
        }

        slow = head;
        while (slow != fast) {
            slow = slow.next;
            fast = fast.next;
        }
        return slow;
    }
}

3. Time & Space Complexity

时间复杂度O(n), 空间复杂度O(1)