找出两个链表中合并的元素

function ListNode(val) {    this.val = val    this.next = null}
/** * @param {Integer[]} arr * @return {ListNode} */function createList(arr) {    if (arr.length === 0) {        return null    }    let root = new ListNode(arr[0])    let curtNode = root    for (let i = 1; i < arr.length; i++) {        let nextNode = new ListNode(arr[i])        curtNode.next = nextNode        curtNode = nextNode    }    return root}
/** * @param {ListNode} headA * @param {ListNode} headB * @return {ListNode} */function getListLength(node) {    let len = 0    let curtNode = node    while (curtNode) {        curtNode = curtNode.next        len += 1    }    return len}
function getIntersectionNode(headA, headB) {    // Get list A and list B length: m, n    let m = getListLength(headA)    let n = getListLength(headB)    let nodeA = headA    let nodeB = headB    // Have two pointers: p1 for A, p2 for B    // - if m >= n: p1 = m - n, p2 = 0    // - else: p1 = 0, p2 = n - m    if (m >= n) {        for (let i = 0; i < m - n; i++) {            nodeA = nodeA.next        }    } else {        for (let i = 0; i < n - m; i++) {            nodeB = nodeB.next        }    }    // Increment p1 and p2 in the same time and compare, exit when node(p1).val === node(p2).val    while (nodeA) {        if (nodeA.val === nodeB.val) {            return nodeA        }        nodeA = nodeA.next        nodeB = nodeB.next    }    return null}
function test1() {    let headA = createList([4, 1, 8, 4, 5])    let headB = createList([7, 10, 5, 6, 11, 8, 4, 5])    console.log(getListLength(headA))    console.log(getListLength(headB))    let node = getIntersectionNode(headA, headB)    console.log(node.val)}
test1()