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141. Linked List Cycle
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?
思路分析
除了双指针外，还可以使用 Map 来解决。只是空间复杂度要高一些。
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/**
* Runtime: 0 ms, faster than 100.00% of Java online submissions for Linked List Cycle.
*
* Memory Usage: 37.9 MB, less than 84.29% of Java online submissions for Linked List Cycle.
*/
public boolean hasCycle(ListNode head) {
if (Objects.isNull(head)) {
return false;
}
ListNode slow = head;
ListNode fast = head.next;
while (Objects.nonNull(fast)) {
if (slow == fast) {
return true;
}
slow = slow.next;
fast = fast.next;
if (Objects.nonNull(fast)) {
fast = fast.next;
}
}
return false;
}
/**
* Runtime: 5 ms, faster than 6.60% of Java online submissions for Linked List Cycle.
*
* Memory Usage: 38.3 MB, less than 82.14% of Java online submissions for Linked List Cycle.
*
* Copy from: https://leetcode.com/problems/linked-list-cycle/solution/[Linked List Cycle solution - LeetCode]
*/
public boolean hasCycleMap(ListNode head) {
Set<ListNode> nodes = new HashSet<>();
ListNode current = head;
while (Objects.nonNull(current)) {
if (nodes.contains(current)) {
return true;
} else {
nodes.add(current);
}
current = current.next;
}
return false;
}
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/**
* 快慢指针
*/
public boolean hasCycle(ListNode head) {
if (head == null || head.next == null) {
return false;
}
ListNode slow = head;
ListNode fast = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
if (slow == fast) {
return true;
}
}
return false;
}