LeetCode 141. Linked List Cycle环形链表 (C++)

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;
    }
};