Added 2. Add Two Numbers

Author: BaileyKau

Diff for: 2. Add Two Numbers.js

@@ -0,0 +1,78 @@
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.next = (next===undefined ? null : next)
+ * }
+ */
+
+/**
+ * @param {ListNode} l1
+ * @param {ListNode} l2
+ * @return {ListNode}
+ */
+
+/**
+    Runtime: 136 ms, faster than 61.77% of JavaScript online submissions for Add Two Numbers.
+    Memory Usage: 46.8 MB, less than 15.00% of JavaScript online submissions for Add Two Numbers.
+ */
+
+var addTwoNumbers = function(l1, l2) {
+    let head = new ListNode();
+        let curr = head;
+        let extra = 0;
+        while (l1 || l2) {
+            let l1Val = l1 ? l1.val : 0;
+            let l2Val = l2 ? l2.val : 0;
+            let sum = l1Val + l2Val + extra
+            extra = sum > 9 ? 1 : 0;
+            if (l1) {
+                l1.val = sum % 10;
+                curr.next = l1;
+            } else {
+                l2.val = sum % 10;
+                curr.next = l2;
+            }
+            curr = curr.next;
+            if (l1) l1 = l1.next;
+            if (l2) l2 = l2.next;
+        }
+        if (extra > 0) {
+            curr.next = new ListNode(1);
+        }
+        return head.next;
+    };
+
+/** First Try
+ * - Didn't work because logic flawed
+ * - Input: [1, 0, 0 ,1], [3, 4]
+ * - Expected Output: [1, 0, 3, 5]
+ * - Actual Output: [4, 4, 0, 1] 
+ */
+
+var addTwoNumbers = function(l1, l2) {
+    let sum = 0;
+    let m = 0;
+    let runner = l1;
+    while (runner) {
+        sum += (runner.val * Math.pow(10, m));
+        runner = runner.next;
+        m++;
+    }
+    console.log(sum);
+    runner = l2, m = 0;
+    while (runner) {
+        sum += (runner.val * Math.pow(10, m));
+        runner = runner.next;
+        m++;
+    }
+    sum = sum.toString().split('').reverse();
+    let l3 = new ListNode(sum[0], null);
+    runner = l3;
+    for (let i = 1; i < sum.length; i++) {
+        runner.next = new ListNode(sum[i], null);
+        runner = runner.next;
+    }
+    return l3;
+};
+