diff --git a/0023.Merge-k-Sorted-Lists/memo.md b/0023.Merge-k-Sorted-Lists/memo.md
new file mode 100644
index 0000000..74f9dc9
--- /dev/null
+++ b/0023.Merge-k-Sorted-Lists/memo.md
@@ -0,0 +1,108 @@
+# 23. Merge k Sorted Lists
+
+## step1
+
+17mで解けた。紙に書いて考えるうちに最小値を管理する -> heapという発想になった。破壊的な解法になっている。
+
+時間計算量：
+O(sum(lists[i].length)log k)
+
+空間計算量：
+O(k)
+
+## step2
+
+非破壊的な解法: step2.py
+
+indexを格納しないと、<が定義されずエラーになる。
+
+これを面接のtest環境のないところで気がつくのは、ヒントがないと自分には難しそう。
+
+以下の実験でも確認できる。
+
+```python
+heap = [(0, ListNode()), (0, ListNode())]
+heapq.heapify(heap)
+# TypeError: '<' not supported between instances of 'ListNode' and 'ListNode'
+```
+
+### 追記
+ListNodeに__lt__メソッドを強制的に加えた実装を追加
+
+## 他の人のコード
+
+https://github.com/shining-ai/leetcode/pull/67
+
+> これ (index) がないと、定義されていないListNodeの比較になって問題が出るのですかね。
+
+- 非破壊的+heapを使わない解法：
+
+```python
+lass Solution:
+    def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:
+        lists_val = []
+        for node in lists:
+            while node:
+                lists_val.append(node.val)
+                node = node.next
+        lists_val.sort()
+        sentinel = ListNode()
+        node = sentinel
+        for i in lists_val:
+            node.next = ListNode(i)
+            node = node.next
+
+        return sentinel.next
+```
+
+LinkedListの操作に気を取られて思いつかなかった。
+
+時間計算量的はS=sum(lists[i].length)として O(Slog S)だが実際に実行してみると、heapの解法と大差なかった。
+
+
+> マージソートをイメージした解法
+> 先頭の2つのリストをマージしていき、最後の1つになるまで繰り返す
+
+自然な解法なので、面接で聞かれる可能性は高そう。これを思いつかなったのは、まだまだ修行が足りないということだろうな。
+
+時間計算量 O(Slog k)
+
+sentinelという変数名は良いな
+
+```python
+
+class Solution:
+    def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:
+        def merge_two_lists(list_1, list_2):
+            sentinel = ListNode()
+            node = sentinel
+            while list_1 and list_2:
+                if list_1.val < list_2.val:
+                    node.next = list_1
+                    list_1 = list_1.next
+                else:
+                    node.next = list_2
+                    list_2 = list_2.next
+                node = node.next
+            if not list_1:
+                node.next = list_2
+            else:
+                node.next = list_1
+            return sentinel.next
+
+        list_queue = deque(lists)
+        if not list_queue:
+            return None
+        while 1:
+            list_1 = list_queue.popleft()
+            if not list_queue:
+                return list_1
+            list_2 = list_queue.popleft()
+            mearged_list = merge_two_lists(list_1, list_2)
+            list_queue.append(mearged_list)
+```
+
+## step3
+マージソートの解法を3回書くことにする。
+
+merge_two_listsの内側でswapを用いることでsimpleになった気がする。
diff --git a/0023.Merge-k-Sorted-Lists/step1_in_place.py b/0023.Merge-k-Sorted-Lists/step1_in_place.py
new file mode 100644
index 0000000..89a0246
--- /dev/null
+++ b/0023.Merge-k-Sorted-Lists/step1_in_place.py
@@ -0,0 +1,29 @@
+import heapq
+
+
+# Definition for singly-linked list.
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
+
+class Solution:
+    def mergeKLists(self, lists: list[ListNode | None]) -> ListNode | None:
+        heap_indices = [
+            (node.val, i) for i, node in enumerate(lists) if node is not None
+        ]
+        heapq.heapify(heap_indices)
+        dummy = ListNode()
+
+        node = dummy
+        while heap_indices:
+            _, i = heapq.heappop(heap_indices)
+            next_node = lists[i]
+            node.next = next_node
+            lists[i] = next_node.next
+            if next_node.next is not None:
+                heapq.heappush(heap_indices, (next_node.next.val, i))
+            node = node.next
+
+        return dummy.next
diff --git a/0023.Merge-k-Sorted-Lists/step2_not_in_place.py b/0023.Merge-k-Sorted-Lists/step2_not_in_place.py
new file mode 100644
index 0000000..38f3996
--- /dev/null
+++ b/0023.Merge-k-Sorted-Lists/step2_not_in_place.py
@@ -0,0 +1,26 @@
+import heapq
+
+
+# Definition for singly-linked list.
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
+
+class Solution:
+    def mergeKLists(self, lists: list[ListNode | None]) -> ListNode | None:
+        # Include index i to avoid comparison collisions when node values are equal.
+        heap = [(node.val, i, node) for i, node in enumerate(lists) if node is not None]
+        heapq.heapify(heap)
+
+        dummy = ListNode()
+        node = dummy
+        while heap:
+            val, i, head = heapq.heappop(heap)
+            node.next = ListNode(val)
+            node = node.next
+            if head.next is not None:
+                heapq.heappush(heap, (head.next.val, i, head.next))
+
+        return dummy.next
diff --git a/0023.Merge-k-Sorted-Lists/step2_not_in_place_lt.py b/0023.Merge-k-Sorted-Lists/step2_not_in_place_lt.py
new file mode 100644
index 0000000..c09f6b4
--- /dev/null
+++ b/0023.Merge-k-Sorted-Lists/step2_not_in_place_lt.py
@@ -0,0 +1,27 @@
+import heapq
+
+
+# Definition for singly-linked list.
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
+
+class Solution:
+    def mergeKLists(self, lists: list[ListNode | None]) -> ListNode | None:
+        ListNode.__lt__ = lambda self, other: self.val < other.val
+
+        heap = [node for node in lists if node is not None]
+        heapq.heapify(heap)
+
+        dummy = ListNode()
+        node = dummy
+        while heap:
+            head = heapq.heappop(heap)
+            node.next = ListNode(head.val)
+            node = node.next
+            if head.next is not None:
+                heapq.heappush(heap, head.next)
+
+        return dummy.next
diff --git a/0023.Merge-k-Sorted-Lists/step3_merge.py b/0023.Merge-k-Sorted-Lists/step3_merge.py
new file mode 100644
index 0000000..e1a7cae
--- /dev/null
+++ b/0023.Merge-k-Sorted-Lists/step3_merge.py
@@ -0,0 +1,38 @@
+# Definition for singly-linked list.
+import collections
+
+
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
+
+class Solution:
+    def mergeKLists(self, lists: list[ListNode | None]) -> ListNode | None:
+        if not lists:
+            return None
+
+        def merge_two_lists(node1, node2):
+            sentinel = ListNode()
+            node = sentinel
+
+            while node1 is not None and node2 is not None:
+                if node1.val > node2.val:
+                    node1, node2 = node2, node1
+                node.next = node1
+                node = node.next
+                node1 = node1.next
+
+            node.next = node1 if node1 is not None else node2
+            return sentinel.next
+
+        node_to_merge = collections.deque(lists)
+        while 1:
+            node1 = node_to_merge.popleft()
+
+            if not node_to_merge:
+                return node1
+
+            node2 = node_to_merge.popleft()
+            node_to_merge.append(merge_two_lists(node1, node2))