141 . Linked List Cycle
Easy
Given a linked list, determine if it has a cycle in it.
Follow up:
Can you solve it without using extra space?
1ms:
public boolean hasCycle(ListNode head) {
if(head==null) return false;
ListNode slow = head;
ListNode fast = head;
while(fast.next!=null&&fast.next.next!=null){
slow = slow.next;
fast = fast.next.next;
if(slow==fast)
return true;
}
return false;
}
142 . Linked List Cycle II
Medium
Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
Note: Do not modify the linked list.
Follow up:
Can you solve it without using extra space?
本文介绍两种链表循环检测算法:一种用于判断链表是否存在循环,另一种用于找出循环开始的节点。文章提供了简洁高效的代码实现,并探讨了如何在不使用额外空间的情况下解决这两个问题。
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