C++怎么解决单链表中的环问题

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单链表中的环

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.

Follow up:

Can you solve it using O(1) (i.e. constant) memory?

这道题是快慢指针的经典应用。只需要设两个指针，一个每次走一步的慢指针和一个每次走两步的快指针，如果链表里有环的话，两个指针最终肯定会相遇。实在是太巧妙了，要是我肯定想不出来。代码如下：

C++ 解法：

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;    }};

Java 解法：

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;    }}

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