動機

複習判圈

Problem

Given head, the head of a linked list, determine if the linked list has a cycle in it.

There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail's next pointer is connected to. Note that pos is not passed as a parameter.

Return true if there is a cycle in the linked list. Otherwise, return false.

 

Example 1:

Input: head = [3,2,0,-4], pos = 1Output: trueExplanation: There is a cycle in the linked list, where the tail connects to the 1st node (0-indexed).

Example 2:

Input: head = [1,2], pos = 0Output: trueExplanation: There is a cycle in the linked list, where the tail connects to the 0th node.

Example 3:

Input: head = [1], pos = -1Output: falseExplanation: There is no cycle in the linked list.

 

Constraints:

  • The number of the nodes in the list is in the range [0, 104].
  • -105 <= Node.val <= 105
  • pos is -1 or a valid index in the linked-list.

 

Follow up: Can you solve it using O(1) (i.e. constant) memory?

Sol

class Solution:
    def hasCycle(self, head: ListNode) -> bool:
        if head is None or head.next is None:
            return False
        a = b = head
        while b is not None and b.next is not None:
            a = a.next
            b = b.next.next
            if a is b:
                break
        return a is b

原本想著怎麼把break拿掉，結果看到以前的code 直接把比較快的加一就ok了…

class Solution {
public:
    bool hasCycle(ListNode *head) {
        if(!head || head->next == NULL)
            return 0;
        ListNode * a = head;
        ListNode * b = head->next;
        for(;(b && b->next) && a != b;a=a->next,b=b->next->next);
        return a == b;
    }
};