二指针算法解释

const findpair = (arr, target) =&gt; {    let left = 0;                 // start with two pointers left from start, right, from the end    let right = arr.length - 1;    while (left <p>让我们切换到指针具有不同速度的方法。这是一个常见的问题，你可以在面试中遇到。您需要找到给定链接列表的中间位置。<br>蛮力方法并不像前面的例子那么糟糕，但面试官期望有更好的方法。<br>使用两个指针算法，您将以 o(n) 复杂度解决此问题，而如果您使用两个顺序循环，则暴力方法将需要 o(2n)。</p><pre class="brush:php;toolbar:false">class ListNode {    constructor(value) {        this.value = value;        this.next = null;    }}const findMiddle = (head) =&gt; {    if (!head) return null;    let slow = head;    let fast = head;    while (fast &amp;&amp; fast.next) {        slow = slow.next;           // Move slow pointer one step        fast = fast.next.next;      // Move fast pointer two steps    }    return slow;  // Slow pointer will be at the middle}// Creating a linked list 1 -&gt; 2 -&gt; 3 -&gt; 4 -&gt; 5const head = new ListNode(1);const node2 = new ListNode(2);const node3 = new ListNode(3);const node4 = new ListNode(4);const node5 = new ListNode(5);head.next = node2;node2.next = node3;node3.next = node4;node4.next = node5;findMiddle(head).value);  // Output: 3 (middle node)