Linked List Cycle (E)
题目
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?
题意
判断给定链表中是否存在环。
思路
比较简单的就是将所有遍历到的结点记录下来，如果记录了两次则说明存在环。
(O(1))空间的方法是使用快慢指针，慢指针每次走一步，快指针每次走两步，如果快指针追上慢指针则说明存在环。实际上可以看做一个相遇问题，如果快慢指针都在一个环中，且快指针距离慢指针还有n个结点的距离，那么经过n个回合后快指针必定与慢指针重合。
代码实现
Java
Hash
public class Solution {
public boolean hasCycle(ListNode head) {
if (head == null) {
return false;
}
Set<ListNode> set = new HashSet<>();
while (head != null) {
if (!set.add(head)) {
return true;
}
head = head.next;
}
return false;
}
}
快慢指针
public class Solution {
public boolean hasCycle(ListNode head) {
if (head == null) {
return false;
}
ListNode slow = head, fast = head;
while (fast.next != null && fast.next.next != null) {
slow = slow.next;
fast = fast.next.next;
if (fast == slow) {
return true;
}
}
return false;
}
}
JavaScript
/**
* @param {ListNode} head
* @return {boolean}
*/
var hasCycle = function (head) {
const nodes = new Set()
while (head) {
if (nodes.has(head)) {
return true
}
nodes.add(head)
head = head.next
}
return false
}