topic
Link: https: //leetcode-cn.com/problems/linked-list-cycle
Given a list, the list is determined whether a ring.
To show the list given ring, we integer pos connected to the end of the list to represent the position of the list (index starts from 0). If pos is -1, no ring on this list.
Example 1:
Input: head = [3,2,0, -4], pos = 1
Output: true
Explanation: the list has a ring tail portion connected to a second node.
Example 2:
Input: head = [1,2], pos = 0
Output: true
Explanation: the list has a ring, which is connected to the tail of the first node.
Example 3:
Input: head = [1], pos = -1
Output: false
Explanation: The list does not ring.
Advanced:
You can use O (1) (ie, constant) memory to solve this problem?
C ++ code
Ideas: Use pointer speed
The mobile node a pointer to a slow
A two node pointer moves fast
Once the fast pointer is NULL, there is no ring, because it can be found at the end of
If there is a ring, then the whole cycle is endless, the pointer will soon overtake slower pointer, and will not be missed.
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
bool hasCycle(ListNode *head) {
if(head == NULL || head->next == NULL)
return false;
ListNode* slow = head;
ListNode* fast = head->next;
while(slow != fast){
if(fast->next == NULL || fast->next->next == NULL)
return false;
fast = fast->next->next;
slow = slow->next;
}
return true;
}
};
When execution: 16ms, defeated 54.04% of all users to submit in C ++
Memory consumption: 9.8MB, defeated 33.30% of all users to submit in C ++
