141. Linked List Cycle

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本文介绍了一种使用快慢指针技巧来判断链表中是否存在循环的有效方法。通过两个速度不同的指针遍历链表，如果存在循环，快指针最终会追上慢指针；若链表结束则不存在循环。

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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.

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.

Example 3:

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

快慢指针，妙啊

参考这里

/**
 * 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 *fast=head,*slow=head;
        while(fast&&slow){
            if(fast->next){
                fast=fast->next->next;
            }
            else{
                return false;
            }
            slow=slow->next;
            if(fast==slow){
                return true;
            }
        }
        return false;
    }
};