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141.linked-list-cycle.cpp
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141.linked-list-cycle.cpp
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/*
* @lc app=leetcode id=141 lang=cpp
*
* [141] Linked List Cycle
*
* https://leetcode.com/problems/linked-list-cycle/description/
*
* algorithms
* Easy (37.60%)
* Likes: 1692
* Dislikes: 200
* Total Accepted: 433.5K
* Total Submissions: 1.2M
* Testcase Example: '[3,2,0,-4]\n1'
*
* 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?
*
*/
/**
* 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)
return false;
if (!head -> next)
return false;
if (!head -> next -> next)
return false;
ListNode* slow = head -> next;
ListNode* fast = head -> next -> next;
while (slow != fast) {
if (!fast -> next)
return false;
if (!fast -> next -> next)
return false;
slow = slow -> next;
fast = fast -> next -> next;
}
return true;
}
};