LeetCode 141. Linked List Cycle

Given a linked list, determine if it has a cycle in it.

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

 

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.

 

Follow up:

Can you solve it using O(1) (i.e. constant) memory?

隊列中找環，其實只管想法就是把它逐個放入Set或者List裏去，然後用contains方法，但其後續想降低空間複雜度，便想到經典的雙指針走法，Accepted代碼如下：

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public boolean hasCycle(ListNode head) {
        if (head == null || head.next == null) {
            return false;
        }
        ListNode p1 = head;
        ListNode p2 = head;
        while (p2 != null && p2.next != null) {
            p1 = p1.next;
            p2 = p2.next.next;
            if (p1.equals(p2)) {
                return true;
            }
        }
        return false;
    }
}

該方法時間和空間複雜度均能超過當前100%的答案