【LeetCode】141. Linked List Cycle[快慢指针]
【LeetCode】141. Linked List Cycle[快慢指针]
Given a linked list, determine if it has a cycle in it.
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.
Example 1:
Input: head = [3,2,0,-4], pos = 1 Output: true 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: true Explanation: There is a cycle in the linked list, where tail connects to the first node.
Example 3:
Input: head = [1], pos = -1 Output: false Explanation: There is no cycle in the linked list.
Follow up:
Can you solve it using O(1) (i.e. constant) memory?
来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/linked-list-cycle 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
跑两步的追跑一步的~
/**
* 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) {
ListNode* fa=head;
ListNode* sl=head;
while(fa&&fa->next){
fa=fa->next;
fa=fa->next;
sl = sl->next;
if(sl==fa){return true;}
}return false;
}
};最快的老哥和我差在哪里？
if(head == nullptr || head->next == nullptr)
return false;
加了这一个判断就往前走了不少20ms变成8ms
/**
* 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 == nullptr || head->next == nullptr)
return false;
ListNode* slow = head->next;
ListNode* fast = slow->next;
while(slow != fast){
if(slow == nullptr || fast == nullptr)
return false;
slow = slow->next;
fast = fast->next;
if(fast != nullptr)
fast = fast->next;
}
return true;
}
};腾讯云开发者
