141 Linked List Cycle 环形链表

Description:
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:

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 1

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 2

Example 3:
Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.

Example 3

Follow up:

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

题目描述:
给定一个链表，判断链表中是否有环。

为了表示给定链表中的环，我们使用整数 pos 来表示链表尾连接到链表中的位置（索引从 0 开始）。 如果 pos 是 -1，则在该链表中没有环。

示例：

示例 1：
输入：
head = [3,2,0,-4], pos = 1
输出：
true
解释：
链表中有一个环，其尾部连接到第二个节点。

示例 1

示例 2：
输入：
head = [1,2], pos = 0
输出：
true
解释：
链表中有一个环，其尾部连接到第一个节点。

示例 2

示例 3：
输入：
head = [1], pos = -1
输出：
false
解释：
链表中没有环。

示例 3

进阶：
你能用 O(1)（即，常量）内存解决此问题吗？

思路:

快慢指针
时间复杂度O(n), 空间复杂度O(1)

代码:
C++:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution 
{
public:
    bool hasCycle(ListNode *head) 
    {
        ListNode *fast = head, *slow = head;
        while (fast and fast -> next) 
        {            
            fast = fast -> next -> next;
            slow = slow -> next;
            if (fast == slow) return true;
        }
        return false;
    }
};

Java:

/**
 * 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) {
        ListNode fast = head, slow = head;
        while (fast != null && fast.next != null) {            
            fast = fast.next.next;
            slow = slow.next;
            if (fast == slow) return true;
        }
        return false;
    }
}

Python:

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution(object):
    def hasCycle(self, head: ListNode) -> ListNode:
        slow = fast = head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if slow == fast:
                return True
        return False