perf: Update addTwoNumbers的JavaScript解法 (azl397985856#163)

* refactor: Update addTwoNumbers的JavaScript解法

* style: 提交的addTwoNumbers的JS代码格式改成2个空格

* chore: Add 2.addTwoNumbers的JS解法的注释

problems/2.addTwoNumbers.md

@@ -47,46 +47,39 @@ JavaScript:
  * @return {ListNode}
  */
 var addTwoNumbers = function(l1, l2) {
-  var carried = 0; // 用于进位
-  const head = new ListNode();
-  const noop = {
-    val: 0,
-    next: null
-  };
-  let currentL1 = l1;
-  let currentL2 = l2;
-  let currentNode = head; // 返回的链表的当前node
-  let newNode; // 声明在外面节省内存
-  let previousNode; // 记录前一个节点，便于删除最后一个节点
-
-  while (currentL1 || currentL2) {
-    newNode = new ListNode(0);
-
-    currentNode.val =
-      ((currentL1 || noop).val + (currentL2 || noop).val + carried) % 10;
-
-    currentNode.next = newNode;
-    previousNode = currentNode;
-    currentNode = newNode;
-
-    if ((currentL1 || noop).val + (currentL2 || noop).val + carried >= 10) {
-      carried = 1;
-    } else {
-      carried = 0;
+  if (l1 === null || l2 === null) return null
+
+  // 使用dummyHead可以简化对链表的处理，dummyHead.next指向新链表
+  let dummyHead = new ListNode(0)
+  let cur1 = l1
+  let cur2 = l2
+  let cur = dummyHead // cur用于计算新链表
+  let carry = 0 // 进位标志
+
+  while (cur1 !== null || cur2 !== null) {
+    let val1 = cur1 !== null ? cur1.val : 0
+    let val2 = cur2 !== null ? cur2.val : 0
+    let sum = val1 + val2 + carry
+    let newNode = new ListNode(sum % 10) // sum%10取模结果范围为0~9，即为当前节点的值
+    carry = sum >= 10 ? 1 : 0 // sum>=10，carry=1，表示有进位
+    cur.next = newNode
+    cur = cur.next
+
+    if (cur1 !== null) {
+      cur1 = cur1.next
     }
 
-    currentL1 = (currentL1 || noop).next;
-    currentL2 = (currentL2 || noop).next;
+    if (cur2 !== null) {
+      cur2 = cur2.next
+    }
   }
 
-  if (carried) {
-    // 还有位没进呢
-    previousNode.next = new ListNode(carried)
-  } else {
-    previousNode.next = null;
+  if (carry > 0) {
+    // 如果最后还有进位，新加一个节点
+    cur.next = new ListNode(carry)
   }
 
-  return head;
+  return dummyHead.next
 };
 ```
 C++