本文主要是介绍[LeetCode] Linked List Cycle II,希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!
Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
Follow up:
Can you solve it without using extra space?
解题思路
设链表长度为n,头结点与循环节点之间的长度为k。定义两个指针slow和fast,slow每次走一步,fast每次走两步。当两个指针相遇时,有:
- fast = slow * 2
- fast - slow = (n - k)的倍数
由上述两个式子可以得到slow为(n-k)的倍数
两个指针相遇后,slow指针回到头结点的位置,fast指针保持在相遇的节点。此时它们距离循环节点的距离都为k,然后以步长为1遍历链表,再次相遇点即为循环节点的位置。
实现代码
/*** Definition for singly-linked list.* struct ListNode {* int val;* ListNode *next;* ListNode(int x) : val(x), next(NULL) {}* };*///Runtime:16 ms
class Solution {
public:ListNode *detectCycle(ListNode *head) {if (head == NULL){return NULL;}ListNode *slow = head;ListNode *fast = head;while (fast->next && fast->next->next){slow = slow->next;fast = fast->next->next;if (fast == slow){break;}}if (fast->next && fast->next->next){slow = head;while (slow != fast){slow = slow->next;fast = fast->next;}return slow;}return NULL;}
};
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