leetcode twopointer 141 circular linked list

THE

problem

achieve

Ideas

Judge the ring and use the fast or slow pointer to go. The fast pointer moves twice at a time, and the slow pointer moves one grid at a time. If there is a ring, then it must meet, and when it meets, it is time to exit.

• define fast , slow
• while(fast!&&fast.next!==NULL)
• • fast = fast +2
• • slow = slow +1
• • if(fast ==slow ) return true
• return false;

Summary and reflection

1. Pay attention to how next.next guarantees that it does not cross the boundary.

Code

class Solution {
    
    
public:
   bool hasCycle(ListNode *head) {
    
    
       ListNode* fast = head;
       ListNode* slow = head;
       while(fast && fast->next){
    
      // make sure next is not null that make next.next is not null. 
           fast = fast->next->next;
           slow = slow->next;
           if(fast == slow ) return true;
       }
       return false;
   }
};