Javascript 为什么我达到了时间限制！LeetCode链接列表循环(已解决，但需要解释！)

我正在解决LeetCode问题141. Linked List Cycle:

给定head(linked list的头部)，确定该linked list中是否有一个循环.

如果列表中有某个可以通过连续跟随next指针再次到达的 node ，则该列表中存在一个循环.在内部，pos用于表示tail的next指针所连接的 node 的索引.Note that 101 is not passed as a parameter.

返回true if there is a cycle in the linked list.否则，返回false.

例1:

Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where the tail connects to the 1^st node (0-indexed).

我的原始解决方案(版本A)超出了上述输入的时间限制.我最终能够通过更改two lines of code(版本B)足够快地制定解决方案，解决方案被接受并且在某种程度上更高效.

为什么Version B比Version A快？为什么变量分配比使用原始对象更快/更有效？

我改变了这个.

版本 A ( 失效溶液 )

/**
 * @param {ListNode} head
 * @return {boolean}
 */
var hasCycle = function(head) {
    // create a fast pointer
    let fastPointer = null;
    // create a while loop that lasts for as long as head exists
    while (head) {
        // move fast pointer over by 2
        fastPointer = head?.next?.next || null;
        // if fastpointer is null, this indicates that there is no cycle so return false
        if (fastPointer === null) return false;
        // compare fastPointer to the slow pointer, if they are ever equal to the same node then the list contains a cycle.
        if (fastPointer === head) return true;
        // move head by 1
        head = head.next;
    }
    // if the while loop ends, it indicates the there is no cycle so return false
    return false;
};   

对此.

版本 B ( 工作溶液 )

var hasCycle = function(head) {
    // create a fast pointer
    let fastPointer = head; // !!!CHANGE #1
    // create a while loop that lasts for as long as head exists
    while (head) {
        // move fast pointer over by 2 
        // !!!CHANGE #2 and solution worked!!!!
        fastPointer = fastPointer?.next?.next || null;
        // if fastpointer is null, this indicates that there is no cycle so return false
        if (fastPointer === null) return false;
        // compare fastPointer to the slow pointer, if they are ever equal to the same node then the list contains a cycle.
        if (fastPointer === head) return true;
        // move head by 1
        head = head.next;
    }
    // if the while loop ends, it indicates the there is no cycle so return false
    return false;
};