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.

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;
}
}

Github 同步地址:

https://github.com/grandyang/leetcode/issues/141

类似题目:

Linked List Cycle II

Happy Number

参考资料:

https://leetcode.com/problems/linked-list-cycle/

https://leetcode.com/problems/linked-list-cycle/discuss/44489/O(1)-Space-Solution

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