难度：Easy
题目要求
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?
中文说明
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。
思路1
遍历链表并用哈希表保存遇到的元素。
public class Solution {
public boolean hasCycle(ListNode head) {
Set<ListNode> m = new HashSet<>();
while (head != null) {
if (m.contains(head)) return true;
m.add(head);
head = head.next;
}
return false;
}
}
- 时间复杂度：O(N)
- 空间复杂度：O(N)
思路2
通过两个快慢指针slow和fast，其中slow每次步进一个，fast每次步进两个。 当链表有环的话，slow和fast会相遇。
public class Solution {
public boolean hasCycle(ListNode head) {
if (head == null) return false;
ListNode slow = head, fast = head.next;
while (fast != null) {
if (slow == fast) return true;
slow = slow.next;
fast = fast.next;
if (fast != null) fast = fast.next;
}
return false;
}
}
更优雅的代码实现：
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;
}
}
- 时间复杂度：O(N)
- 空间复杂度：O(1)
