LeetCode 142 Linked List Cycle II

142 Linked List Cycle II

https://leetcode.com/problems/linked-list-cycle-ii/

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail’s next pointer is connected to. Note that pos is not passed as a parameter.

Notice that you should 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.

Constraints:

• The number of the nodes in the list is in the range [0, 104].
• -105 <= Node.val <= 105
• pos is -1 or a valid index in the linked-list.

Two Pointers

采用 Floyd Cycle Detection 算法，龟兔赛跑双指针的思想，龟走一步兔走两步，相遇时停下；一方从 head 重新出发，同时一步一步前进，再次相遇时即为环的起始点。

因为链表涉及到空指针，所以需要先提前判断是否为空，防止运行报错。

时间复杂度 O(n)，空间复杂度 O(1)

/**
 * 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) {
        ListNode *pos = NULL, *tortoise = head, *hare = head;
        if(head == NULL || head->next == NULL)
            return NULL;
        while(tortoise->next && hare->next){
            tortoise = tortoise->next;
            hare = hare->next;
            if(tortoise == NULL || hare == NULL)
                return NULL;
            hare = hare->next;
            if(hare == NULL)
                return NULL;
            if(hare && tortoise && hare == tortoise)
                break;
        }
        
        tortoise = head;
        while(tortoise != hare){
            if(tortoise == NULL || hare == NULL)
                return NULL;
            tortoise = tortoise->next;
            hare = hare->next;
        }
        
        return hare;
    }
};

Runtime: 8 ms, faster than 81.59% of C++ online submissions for Linked List Cycle II.

Memory Usage: 7.5 MB, less than 88.27% of C++ online submissions for Linked List Cycle II.

代码简化后如下

/**
 * 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 *hare = head, *tortoise = head;
        while(hare->next && hare->next->next){
            tortoise = tortoise->next;
            hare = hare->next->next;
            if(hare == tortoise){
                tortoise = head;
                while(tortoise != hare){
                    tortoise = tortoise->next;
                    hare = hare->next;
                }
                return hare;
            }
        }
        return NULL;
    }
};

Runtime: 8 ms, faster than 81.59% of C++ online submissions for Linked List Cycle II.

Memory Usage: 7.6 MB, less than 69.29% of C++ online submissions for Linked List Cycle II.