# 题目
https://leetcode-cn.com/problems/linked-list-cycle/
# 思路
快慢指针,慢指针slow每次走一步,快指针fast每次走两步,如果相遇就说明有环,如果fast或者fast->next为空说明没有环。
# 代码
/**
* 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;
ListNode* fast = head->next;
while(slow!=fast){
if(fast==nullptr || fast->next== nullptr){
return false;
}
slow = slow->next;
fast = fast->next->next;
}
return true;
}
};
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