Given a linked list, determine if it has a cycle in it.
Follow up:
Can you solve it without using extra space?
解题思路:分别设置两个指针one和two,遍历的时候,one=one->next,two=two->next->next,这样以不同速率进行遍历,如果该列表存在环的话,one和two必然在某个时刻相等,反之,则可以正常遍历完列表,不会出现两个指针相等的情况。
代码如下:
/**
* 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)
return false;
ListNode* cur1=head,*cur2=head;
while(cur1->next && cur2->next && cur2->next->next)
{
cur1 = cur1->next;
cur2 = cur2->next->next;
if(cur1 == cur2)
return true;
}
return false;
}
};