Linked List Cycle II
Desicription
Given a linked list, return the node where the cycle begins. If there is no cycle, return null
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Note: Do not modify the linked list.
Follow up:
Can you solve it without using extra space?
Solution
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class Solution { public: ListNode* detectCycle(ListNode* head) { ListNode* slow = head; ListNode* fast = head; ListNode* entry = head; while(fast && fast->next && fast->next->next) { slow = slow->next; fast = fast->next->next; if(slow == fast) { while(slow != entry) { slow = slow->next; entry = entry->next; } return entry; } } return NULL; } };
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