141. Linked List Cycle
Given a linked list, determine if it has a cycle in it.
Follow up:
Can you solve it without using extra space?
解法
快慢指针法,当两指针重合,则有环。
/**
* 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) {
if (head == null || head.next == null) {
return false;
}
ListNode slow = head, fast = head.next;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
if (slow == fast) {
return true;
}
}
return false;
}
}
本文介绍了一种使用快慢指针法来判断链表中是否存在环的有效算法。通过两个速度不同的指针遍历链表,如果存在环则两指针最终会相遇。
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