diff --git a/lectures/12_advanced_recursion/README.md b/lectures/12_advanced_recursion/README.md
index b5bbee2..c8f19e5 100644
--- a/lectures/12_advanced_recursion/README.md
+++ b/lectures/12_advanced_recursion/README.md
@@ -94,7 +94,12 @@ into the scratch vector. Finally, the entire scratch vector is copied back.
 - Can we analyze this algorithm and determine the order notation for the number of operations it will perform?
 Count the number of pairwise comparisons that are required.
 
-## 12.6 Example: Word Search
+## 12.6 Merge Sort Exercises
+
+- [Leetcode problem 912: Sort an Array](https://leetcode.com/problems/sort-an-array/). Solution: [p1472_browserhistory.cpp](../../leetcode/p1472_browserhistory.cpp)
+- [Leetcode problem 148: Sort List](https://leetcode.com/problems/sort-list/). Solution: [p148_sortlist.cpp](../../leetcode/p148_sortlist.cpp)
+
+## 12.7 Example: Word Search
 
 - Take a look at the following grid of characters.
 ```console
@@ -117,11 +122,11 @@ daldruetryrt
   – At each such location, the occurrences of the second letter are sought in the 4 locations surrounding (r, c).
   - At each location where the second letter is found, a search is initiated in the 4 locations surrouding this second letter.
 
-## 12.7 Exercise: Complete the implementation
+## 12.8 Exercise: Complete the implementation
 
 - [Leetcode problem 79: Word Search](https://leetcode.com/problems/word-search/). Solution: [p1472_browserhistory.cpp](../../leetcode/p1472_browserhistory.cpp)
 
-## 12.8 Summary of Nonlinear Word Search Recursion
+## 12.9 Summary of Nonlinear Word Search Recursion
 
 - Recursion starts at each location where the first letter is found.
 - Each recursive call attempts to find the next letter by searching around the current position. When it is found,
@@ -129,8 +134,6 @@ a recursive call is made.
 - The current path is maintained at all steps of the recursion.
 - The “base case” occurs when the path is full or all positions around the current position have been tried.
 
-## 12.9 Leetcode Exercises
+## 12.10 Leetcode Exercises
 
 - [Leetcode problem 704: Binary Search](https://leetcode.com/problems/binary-search/). Solution: [p141_linkedlistcycle.cpp](../../leetcode/p141_linkedlistcycle.cpp)
-- [Leetcode problem 912: Sort an Array](https://leetcode.com/problems/sort-an-array/). Solution: [p1472_browserhistory.cpp](../../leetcode/p1472_browserhistory.cpp)
-- [Leetcode problem 148: Sort List](https://leetcode.com/problems/sort-list/). Solution: [p1472_browserhistory.cpp](../../leetcode/p1472_browserhistory.cpp)
diff --git a/leetcode/p148_sortlist.cpp b/leetcode/p148_sortlist.cpp
new file mode 100644
index 0000000..727f113
--- /dev/null
+++ b/leetcode/p148_sortlist.cpp
@@ -0,0 +1,76 @@
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode() : val(0), next(nullptr) {}
+ *     ListNode(int x) : val(x), next(nullptr) {}
+ *     ListNode(int x, ListNode *next) : val(x), next(next) {}
+ * };
+ */
+class Solution {
+public:
+    ListNode* mergeLists(ListNode* head1, ListNode* head2){
+        // the value doesn't matter here
+        ListNode *head = new ListNode(-1);
+        ListNode* current;
+        current = head;
+        if(head1 == nullptr){
+            return head2;
+        }
+        if(head2 == nullptr){
+            return head1;
+        }
+        while(head1!=nullptr && head2!=nullptr){
+            if(head1->val < head2->val){
+                current->next = head1;
+                head1 = head1->next;
+            }else{
+                current->next = head2;
+                head2 = head2->next;
+            }
+            current = current->next;
+        }
+
+        // at this point, one of them has already been merged into the "New" list.
+        if(head1!=nullptr){
+            // this means we still have some elements on the 1st list, now let's merge them into the "New" list.
+                current->next = head1;
+        }else{
+            // this means we still have some elements on the 2nd list, now let's merge them into the "New" list.
+                current->next = head2;
+        }
+        return head->next;
+    }
+
+    ListNode* sortList(ListNode* head) {
+        // base case
+        if(head==nullptr || head->next==nullptr){
+            return head;
+        }
+
+        // split the list
+        ListNode* slow=head;
+        ListNode* fast=head;
+        ListNode* tail1;
+        while(fast!=nullptr && fast->next!=nullptr){
+            tail1 = slow;
+            slow = slow->next;
+            fast = fast->next;
+            fast = fast->next;
+        }
+        // this is the actual split action
+        tail1->next = nullptr;
+        // slow actually is now pointing to the head of the second half
+        // so let's just rename it to make the variable name more meaningful
+        ListNode* head2 = slow;
+
+        // now let's do the recursion
+        head = sortList(head);
+        head2 = sortList(head2);
+
+        // after the two sub-lists are sorted, we merge them into one list
+        head = mergeLists(head, head2);
+        return head;
+    }
+};