Given a linked list, determine if it has a cycle in it.

Follow up:

Can you solve it without using extra space?

给定一个链表，判断是否有环存在。Follow up: 不使用额外空间。

解法：双指针，一个慢指针每次走1步，一个快指针每次走2步的，如果有环的话，两个指针肯定会相遇。

Java：

public class Solution {    public boolean hasCycle(ListNode head) {        ListNode slow = head, fast = head;        while (fast != null && fast.next != null) {            slow = slow.next;            fast = fast.next.next;            if (slow == fast) return true;        }        return false;    }}

Python:

class ListNode:    def __init__(self, x):        self.val = x        self.next = Noneclass Solution:    # @param head, a ListNode    # @return a boolean    def hasCycle(self, head):        fast, slow = head, head        while fast and fast.next:            fast, slow = fast.next.next, slow.next            if fast is slow:                return True        return False　　

C++：

class Solution {public:    bool hasCycle(ListNode *head) {        ListNode *slow = head, *fast = head;        while (fast && fast->next) {            slow = slow->next;            fast = fast->next->next;            if (slow == fast) return true;        }        return false;    }};

　　

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