Javascript 两个索引之间的反向链表

我试图解决这个问题，但我的代码中有一个bug，但我不知道为什么：

给定已排序的链表，删除具有重复编号的所有节点，只保留与原始列表不同的编号

比如说,， 给定1->2->3->3->4->4->5，返回1->2->5。 给定1->1->1->2->3，返回2->3

错误是

Line 44: TypeError: Cannot read property 'val' of null

下面是Javascript中的代码

function ListNode(val) {
    this.val = val;
    this.next = null;
}

var reverseBetween = function(head, m, n) {
    if(m>=n) return head;
    var currentNode = head;
    var previousNode = null;
    var temp = null, tempStart = null;
    var countm = 0, countn=0;
    if(!currentNode.next) return head;
    var startNode = null;
    console.log([head.val, head.next.val, head.next.next.val, head.next.next.next.val, head.next.next.next.next.val]);

    while(countm<m){
        //advance pointer and do nothing
        startNode = previousNode;
        previousNode = currentNode;
        currentNode = currentNode.next;
        countm++;
        console.log(previousNode.val + ', ' + currentNode.val)
    }
    console.log('before: startNode: ' + startNode.val + ', prev: ' + previousNode.val + ', curr: ' +currentNode.val)
    while(countn < (n-m)){

        temp = currentNode.next;
        currentNode.next = previousNode;
        startNode.next = currentNode;
        previousNode.next = temp;


        countn++;
        console.log('startNode: ' + startNode.val + ', prev: ' + previousNode.val + ', curr: ' +currentNode.val)
        console.log([head.val, head.next.val, head.next.next.val, head.next.next.next.val, head.next.next.next.next.val]);

        //advance
        previousNode = currentNode;
        currentNode = currentNode.next;
        console.log('startNode: ' + startNode.val + ', prev: ' + previousNode.val + ', curr: ' +currentNode.val)
    }

    return head;
};

var a = new ListNode(1);
var b = new ListNode(2);
var c = new ListNode(3);
var d = new ListNode(4);
var e = new ListNode(5);
a.next = b;
b.next = c;
c.next = d;
d.next = e;

reverseBetween(a, 2, 4)

函数列表节点（val）{
this.val=val；
this.next=null；
}
var reverseBetween=函数（水头，m，n）{
如果（m>=n）返回头；
var currentNode=头部；
var-previousNode=null；
var temp=null，tempStart=null；
var countm=0，countn=0；
如果（！currentNode.next）返回头；
var startNode=null；
log（[head.val，head.next.val，head.next.next.val，head.next.next.val，head.next.next.val，head.next.next.val]）；
while（countm
函数列表节点（val）{
this.val=val；
this.next=null；
}
var reverseBetween=函数（水头，m，n）{
如果（m>=n）返回头；
var currentNode=头部；
var-previousNode=null；
var temp=null，tempStart=null；
var countm=0，countn=0；
如果（！currentNode.next）返回头；
var startNode=null；
log（[head.val，head.next.val，head.next.next.val，head.next.next.val，head.next.next.val，head.next.next.val]）；
while（count）你能提供一个你使用的链表示例吗？我更新了代码，调用底部的链表@SkylerOK终于得到了这个！我需要保存第二个指针，而不是使用
currentNode
来前进
function ListNode(val) {
    this.val = val;
    this.next = null;
}

var reverseBetween = function(head, m, n) {
    if(m>=n) return head;
    var currentNode = head;
    var previousNode = null;
    var temp = null, tempStart = null;
    var countm = 0, countn=0;
    if(!currentNode.next) return head;
    var startNode = null;
    console.log([head.val, head.next.val, head.next.next.val, head.next.next.next.val, head.next.next.next.next.val]);

    while(countm<m){
        //advance pointer and do nothing
        startNode = previousNode;
        previousNode = currentNode;
        currentNode = currentNode.next;
        countm++;
        console.log(previousNode.val + ', ' + currentNode.val)
    }
    console.log('before: startNode: ' + startNode.val + ', prev: ' + previousNode.val + ', curr: ' +currentNode.val)
    while(countn < (n-m)){

        temp = currentNode.next;
        temp2 = previousNode;
        currentNode.next = previousNode;
        startNode.next = currentNode;
        previousNode.next = temp;


        countn++;
        console.log('startNode: ' + startNode.val + ', prev: ' + previousNode.val + ', curr: ' +currentNode.val)
        console.log([head.val, head.next.val, head.next.next.val, head.next.next.next.val, head.next.next.next.next.val]);

        //advance
        previousNode = temp2;
        currentNode = currentNode.next;
        console.log('startNode: ' + startNode.val + ', prev: ' + previousNode.val + ', curr: ' +currentNode.val)
    }

    return head;
};

var a = new ListNode(1);
var b = new ListNode(2);
var c = new ListNode(3);
var d = new ListNode(4);
var e = new ListNode(5);
a.next = b;
b.next = c;
c.next = d;
d.next = e;

reverseBetween(a, 2, 4)