LeetCode 141. Linked List Cycle--面试编程题--C++,Python解法

题目地址：Linked List Cycle - LeetCode

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.

图见题目网址

Follow up:

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

这道题目是早上**现场面试的编程题。
我当时想到的解法是用set来存所有节点，如果遍历到已经存在的节点，那么这个链表存在环。
Python解法如下：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = Noneclass Solution:def hasCycle(self, head: ListNode) -> bool:s=set()while head!=None:if head in s:return Trues.add(head)head=head.nextreturn False

时间复杂度和空间复杂度都是O(n)。
面试官问我有其他解法吗，我想不出来。。。。。。。。。。。。。

空间复杂度O(1)的做法是使用快慢指针

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = Noneclass Solution:def hasCycle(self, head: ListNode) -> bool:if head is None:return Falseslow, fast = head, head.nextwhile fast is not None and fast.next is not None:slow = slow.nextfast = fast.next.nextif slow == fast:return Truereturn False

C++解法如下：

/*** 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) {if (head == nullptr || head->next == nullptr) {return false;}ListNode *slow = head;ListNode *fast = head->next;while (slow != fast) {if (fast == nullptr || fast->next == nullptr) {return false;}slow = slow->next;fast = fast->next->next;}return true;}
};

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