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 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?
思路:本题要求判断链表中是否存在环。考虑使用两个指针,一个走的快,一个走的慢,如果存在环,则两者必然相遇,返回 true。
C++ 代码
/** * 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 == NULL) return false; ListNode *fast = head; ListNode *slow = head; while(fast->next && fast->next->next) { fast = fast->next->next; slow = slow->next; if(fast == slow) return true; } return false; } };
运行时间:12ms
运行内存:9.7M
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