题目：
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.
分析：
给定一个链表，判断链表中是否有环。
遍历链表将节点存在map中，每次添加时，都判断下，是否已经存在。
还可以用快慢指针，慢指针一次走一个，快指针一次走两个，如果链表存在环的话，快慢指针终会相等。
程序：
/**
* 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) {
map<ListNode*, int> m;
while(head){
if(m[head] == )
return true;
m[head] = ;
head = head->next;
}
return false;
}
};
/**
* 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;
ListNode* slow = head;
while(fast && slow){
fast = fast->next;
slow = slow->next;
if(fast){
fast = fast->next;
}
else
break;
if(fast == slow)
return true;
}
return false;
}
};