[原题链接](https://leetcode-cn.com/problems/middle-of-the-linked-list/solution/qian-duan-shi-tang-ti-jie-chao-hao-li-ji-76rb/) ## 快慢指针 老套路，借助快慢指针，fast 一次走两步，slow 一次走一步，当 fast 到达链表末尾时，slow 就处于链表的中间点了。 ```js const middleNode = function(head) { let fast = head, slow = head; while (fast && fast.next) { slow = slow.next; fast = fast.next.next; } return slow; }; ``` - 时间复杂度: O(n) - 空间复杂度: O(1)
