141. Linked List Cycle

Problem:

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.

思路：
解释一下所给输入中包含pos的意思，因为给的函数里面并没有这个输入。pos存在的意义是为了构造一个有环或无环的链表（从尾节点指向pos指向的节点，为-1则无环），而我们所给的就是头结点head，所以不能用pos来判断是否有环。
1.设置2个指针：fast和slow；
2.fast每次移动2个节点，slow每次移动一个节点；
3.如果fast和slow指向的节点相同，说明有循环，不相同则说明不存在循环。

Solution:

bool hasCycle(ListNode *head) {
    if (!head) return false;
    
    ListNode *fast = head, *slow = head;
    while (fast->next && fast->next->next) {        //这种方法只能证明环存在，而不能找到环开始的节点位置
        fast = fast->next->next;
        slow = slow->next;
        if (fast == slow)   return true;
    }
    return false;
}

性能：
Runtime: 12 ms  Memory Usage: 9.9 MB