LeetCode [链表] 141.Linked List Cycle （C++和Python实现）

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最新推荐文章于 2025-04-30 12:58:09 发布

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该博客介绍了LeetCode中的141题——检测链表中是否存在环。通过使用快慢指针的方法，解释了如何在C++和Python中解决此问题。题目提供了一些示例，并提出了使用O(1)内存的挑战。

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141.Linked List Cycle [难度：简单]

【题目】

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?

【解题C++】

说来惭愧，题目的意思我看了很久都没能看懂。还纠结了好久，为啥不直接用pos的值来判断（当然是因为函数的接口只有链表没有pos啊！！！）嗯，回到正题，这又是一道快慢指针应用的题目。既然是环形，那么快指针就一定有追上慢指针的一天~

/**
 * 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;
        //不要忘了fast->next也不能为空 
        while(fast&&slow&&fast->next)
        {
        	fast = fast->next->next;
        	slow = slow->next;
        	if(slow==fast) return true;
        }
        return false;
    }
};

【题解Python版】

思路同C++。

# 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):
        """
        :type head: ListNode
        :rtype: bool
        """
        fast = slow = head
        while fast and slow and fast.next:
            fast = fast.next.next
            slow = slow.next
            if fast == slow:
                return True
        return False