[算法笔记]-环形链表Ⅱ-解题方法

2024-03-30 15:02:00

原题链接：https://leetcode-cn.com/problems/linked-list-cycle-ii/

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?

引用官方题解的配图：

解法思路：

1. 快慢指针

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode(int x) : val(x), next(NULL) {}* };*/
class Solution {public:ListNode *detectCycle(ListNode *head) {if(!head)return NULL;ListNode *fast = head, *slow = head;bool has_cycle = false;while(fast && fast->next){fast = fast->next->next;slow = slow->next;if(fast == slow){has_cycle = true;break;}}if(has_cycle){ListNode *cycle_start = head;while(cycle_start!=slow){cycle_start = cycle_start->next;slow = slow->next;}return cycle_start;}else{return NULL;}}
};

2. 哈希表

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode(int x) : val(x), next(NULL) {}* };*/
class Solution {public:ListNode *detectCycle(ListNode *head) {unordered_map <ListNode *, int> m;ListNode *p = head;while(p){if(m[p])    return p;m[p]++;p = p->next;}       return NULL;}
};

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