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141. Linked List Cycle
Problem statement
Given a linked list, determine if it has a cycle in it.
Follow up: Can you solve it without using extra space?
Solution
This is a classical problem in a linked list. It is chosen for interview by many high-tech companies.
The generally idea is also a very good philosophy in linked list --> Fast and slow pointers.
- Define two pointers, fast and slow.
- Each time fast goes two step and slow goes one step.
- If there is a loop, they must met sometime.
- Otherwise, fast will become NULL first.
Time complexity is O(n), space complexity is O(1)
The following solution follows the idea to find the middle of a link list. Fast pointer should goes faster than slow at start.
/** * 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->next; ListNode* slow = head; while(fast != NULL && fast->next != NULL){ if(fast == slow){ return true; } fast = fast->next->next; slow = slow->next; } return false; }};
This solution do not need to do the test at beginning of the program, however, fast is initialized the same value of slow.
/** * 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; ListNode* slow = head; while(fast != NULL && fast->next != NULL){ fast = fast->next->next; slow = slow->next; if(fast == slow){ return true; } } return false; }};
141. Linked List Cycle
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