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141. 环形链表 #9

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@Geekhyt

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@Geekhyt

原题链接

链表的基础知识可以移步这个题解,这里不再赘述。

快慢指针

1.使用快慢不同的两个指针遍历,快指针一次走两步,慢指针一次走一步。
2.如果没有环,快指针会先到达尾部,返回 false。
3.如果有环,则一定会相遇。

consthasCycle=function(head){if(!head||!head.next)returnfalse;letfast=head.next;letslow=head;while(fast!==slow){if(!fast||!fast.next){returnfalse;}fast=fast.next.next;slow=slow.next;}returntrue;};
  • 时间复杂度:O(n)
  • 空间复杂度:O(1)

标记法

遍历链表,通过 flag 标记判断是否有环,如果标记存在则有环。(走过的地方插个旗子做标记)

consthasCycle=function(head){while(head){if(head.flag){returntrue;}else{head.flag=true;head=head.next;}}returnfalse;}
  • 时间复杂度:O(n)
  • 空间复杂度:O(1)

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