141 Linked List Cycle 环形链表
Description:
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:
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?
题目描述:
给定一个链表,判断链表中是否有环。
为了表示给定链表中的环,我们使用整数 pos
来表示链表尾连接到链表中的位置(索引从 0 开始)。 如果 pos
是 -1
,则在该链表中没有环。
示例:
示例 1:
输入:
head = [3,2,0,-4], pos = 1
输出:
true
解释:
链表中有一个环,其尾部连接到第二个节点。
示例 2:
输入:
head = [1,2], pos = 0
输出:
true
解释:
链表中有一个环,其尾部连接到第一个节点。
示例 3:
输入:
head = [1], pos = -1
输出:
false
解释:
链表中没有环。
进阶:
你能用 O(1)(即,常量)内存解决此问题吗?
思路:
快慢指针
时间复杂度O(n), 空间复杂度O(1)
代码:
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)
{
ListNode *fast = head, *slow = head;
while (fast and fast -> next)
{
fast = fast -> next -> next;
slow = slow -> next;
if (fast == slow) return true;
}
return false;
}
};
Java:
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public boolean hasCycle(ListNode head) {
ListNode fast = head, slow = head;
while (fast != null && fast.next != null) {
fast = fast.next.next;
slow = slow.next;
if (fast == slow) return true;
}
return false;
}
}
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: ListNode) -> ListNode:
slow = fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
if slow == fast:
return True
return False