Given a linked list, determine if it has a cycle in it. 給定一個連結串列，判斷它是否有環

Follow up: Can you solve it without using extra space?

Solution：

Python

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def hasCycle(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        fast = head
        slow = head
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next
            if fast == slow:
                return True         
        return False

 

C++

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

其實這道題的程式碼很簡單，關鍵在於理解這個程式碼的解法。 要求連結串列是否有環，需要設定兩個指標，一個快指標fast一個慢指標slow，fast每次走兩步，slow每次走一步，如果連結串列有環的話，fast和slow必然會在連結串列環上相遇，這個好比時鐘的秒針和分針。並且通過證明可以知道fast和slow相遇的時，slow必然沒有走完環的一圈(極限情況是slow剛好走完一圈到達環的入口處)。理解了這個原理之後就可以編寫程式碼了。