Update and rename index.js to addTwoNumbers.md

Author: ps0305

leetcode/add-two-numbers/addTwoNumbers.md

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+Approach : Elementary Math
+
+<img width="569" alt="Screenshot 2022-04-08 at 7 32 33 PM" src="https://user-images.githubusercontent.com/34129569/162451709-5de2bf3e-fa18-4025-946d-6b836cfad897.png">
+
+Algorithm
+
+Just like how you would sum two numbers on a piece of paper, 
+we begin by summing the least-significant digits, which is the head of l1 and l2. Since each digit is in the range of 0 \ldots 90…9, summing two digits may "overflow". For example 5 + 7 = 125+7=12. In this case, we set the current digit to 22 and bring over the carry = 1carry=1 to the next iteration. carrycarry must be either 00 or 11 because the largest possible sum of two digits (including the carry) is 9 + 9 + 1 = 199+9+1=19.
+
+The pseudocode is as following:
+
+* Initialize current node to dummy head of the returning list.
+* Initialize carry to 0.
+* Initialize p1 and p2 to head of l1 and l2 respectively.
+* Loop through lists l1l1 and l2l2 until you reach both ends.
+* Set x to node pp's value. If p1 has reached the end of l1, set to 0.
+* Set yy to node qq's value. If qq has reached the end of l2, set to 0.
+* Set sum = x + y + carry.
+* Update carry = sum / 10.
+* Create a new node with the digit value of (sum mod 10) and set it to current node's next, then advance current node to next.
+* Advance both p1 and p2.
+* Check if carry = 1, if so append a new node with digit 11 to the returning list.
+* Return dummy head's next node.
+
+```js
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
+ * }
+ */
+/**
+ * @param {ListNode} l1
+ * @param {ListNode} l2
+ * @return {ListNode}
+ */
+var addTwoNumbers = function(l1, l2) {
+  const dummy = new ListNode();
+  let ptr = dummy;
+  let p1 = l1;
+  let p2 = l2;
+  let c = 0;
+  while (p1 || p2 || c) {
+    const sum = (p1 ? p1.val : 0) + (p2 ? p2.val : 0) + c;
+    c = Math.floor(sum / 10);
+    ptr.next = new ListNode(sum % 10);
+    ptr = ptr.next;
+    p1 = p1 ? p1.next : p1;
+    p2 = p2 ? p2.next : p2;
+  }
+  return dummy.next;
+};