Add merge-two-sorted-lists solution in TypeScript

Author: Jeehay28

merge-two-sorted-lists/Jeehay28.ts

@@ -0,0 +1,82 @@
+// Approach 2:
+// ✅ Time Complexity: O(n)
+// ✅ Space Complexity: O(1)
+
+class ListNode {
+  val: number;
+  next: ListNode | null;
+  constructor(val?: number, next?: ListNode | null) {
+    this.val = val === undefined ? 0 : val;
+    this.next = next === undefined ? null : next;
+  }
+}
+
+function mergeTwoLists(
+  list1: ListNode | null,
+  list2: ListNode | null
+): ListNode | null {
+  let dummy = new ListNode(0);
+  let current = dummy;
+
+  while (list1 && list2) {
+    if (list1.val <= list2.val) {
+      current.next = list1;
+      list1 = list1.next;
+    } else {
+      current.next = list2;
+      list2 = list2.next;
+    }
+    current = current.next;
+  }
+
+  current.next = list1 || list2; // Attach whatever is left
+
+  return dummy.next;
+}
+
+
+// Approach 1: works, but not efficient for big inputs
+// Time Complexity: O(n log n)
+// Space Complexity: O(n)
+
+// function mergeTwoLists(
+//   list1: ListNode | null,
+//   list2: ListNode | null
+// ): ListNode | null {
+//   let stack: number[] = [];
+
+//   const dfs = (node: ListNode | null) => {
+//     if (!node) {
+//       return;
+//     }
+
+//     stack.push(node.val);
+
+//     return dfs(node.next);
+//   };
+
+//   dfs(list1);
+//   dfs(list2);
+
+//   stack.sort((a, b) => a - b);
+
+//   if (stack.length === 0) {
+//     return null;
+//   }
+
+//   let merged = new ListNode();
+//   let dummy = merged;
+
+//   for (let i = 0; i < stack.length; i++) {
+//     dummy.val = stack[i];
+
+//     if (i !== stack.length - 1) {
+//       dummy.next = new ListNode();
+//       dummy = dummy.next;
+//     } else {
+//       dummy.next = null;
+//     }
+//   }
+
+//   return merged;
+// }