141. Linked List Cycle

problem description

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.

1

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.

2

Example 3:

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

3

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

algorithm thought

这里直接用著名的Floyd判圈算法就行，定义一个快指针一个慢指针，快指针每次走两步，慢指针每次走一步。如果有环两个指针肯定能碰到，如果快指针走到NULL节点，那就没有环

code

/**
 * 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* slow = head;
        ListNode* fast = head;

        while(fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;

            if(slow == fast) {
                return true;
            }
        }

        return false;
    }
};

algorithm analysis

快指针每次走两步，如果有环，不用n时间两个节点就能碰到。时间复杂度O(n)