标签:head 142 linked list List fast next Linked cycle
算法描述:
Given a linked list, return the node where the cycle begins. If there is no cycle, return null
.
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.
Note: Do not modify the linked list.
Example 1:
Input: head = [3,2,0,-4], pos = 1 Output: tail connects to node index 1 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: tail connects to node index 0 Explanation: There is a cycle in the linked list, where tail connects to the first node.
Example 3:
Input: head = [1], pos = -1 Output: no cycle Explanation: There is no cycle in the linked list.
Follow up:
Can you solve it without using extra space?
解题思路:首先用快慢两个指针判断是否有环。如果有环,再设一额外指针从头开始,以相同的速度移动,直到与慢指针相遇。此时慢指针指向的位置为环所在位置。注意边界值的问题。
ListNode *detectCycle(ListNode *head) { if(head == nullptr || head->next == nullptr) return nullptr; ListNode* fast = head; ListNode* slow = head; bool hasCycle = false; while(fast->next!=nullptr && fast->next->next!=nullptr){ fast=fast->next->next; slow=slow->next; if(fast == slow){ hasCycle = true; break; } } if(!hasCycle) return nullptr; fast = head; while(fast!=slow){ fast=fast->next; slow=slow->next; } return fast; }
标签:head,142,linked,list,List,fast,next,Linked,cycle 来源: https://www.cnblogs.com/nobodywang/p/10353808.html
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