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+## Step1
+1.
+重複をsetを用いて判定する.
+重複がある場合, previousとcurrentのポインタをずらす. 
+
+```py
+class Solution:
+    def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        seen = set()
+        previous = None
+        current = head
+        while current is not None:
+            if current.val in seen:
+                previous.next = current.next
+                current = current.next
+                continue
+            
+            seen.add(current.val)
+            previous = current
+            current = current.next
+        
+        return head
+```
+2.
+空間計算量がO(1)の解法.
+昇順に並んでいるので, 1つ後ろのノードとのみ比較すればいい.
+
+```py
+class Solution:
+    def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        if head is None:
+            return None
+
+        node = head
+        while node is not None and node.next is not None:
+            if node.val == node.next.val:
+                # next nodeと値が同じ時はnodeを進めない.
+                node.next = node.next.next
+                continue
+
+            node = node.next
+        
+        return head
+```
+
+## Step2
+https://discord.com/channels/1084280443945353267/1195700948786491403/1196399353116499970
+このような解き方がある. Step1の二個目の解法を二重のwhileに書き直したもの.
+
+
+```py
+class Solution:
+    def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        node = head
+        while node is not None:
+            while node.next is not None and node.val == node.next.val:
+                node.next = node.next.next
+
+            node = node.next
+        
+        return head
+```
+また, 
+https://github.com/docto-rin/leetcode/pull/3
+上記を参考に, while => 再帰に書き換えてみる.
+whileの中身をそのままマッピングしていく感じ.
+```py
+class Solution:
+    def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        if head is None or head.next is None:
+            return head
+        
+        if head.val == head.next.val:
+            head.next = head.next.next
+            return self.deleteDuplicates(head)
+        
+        head.next = self.deleteDuplicates(head.next)
+        return head
+
+```
+こう書くこともできる.
+```py
+class Solution:
+    def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        if head is None or head.next is None:
+            return head
+        
+        # head.nextに入るのは, それ以降を再帰的に処理したノードの列.
+        head.next = self.deleteDuplicates(head.next)
+        if head.val == head.next.val:
+            # 先頭を飛ばす.
+            return head.next
+        
+        # 飛ばさない.
+        return head
+```
+
+
+## Step3
+苦手な再帰で練習.
+
+```py
+class Solution:
+    def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        if head is None or head.next is None:
+            return head
+        
+        if head.val == head.next.val:
+            head.next = head.next.next
+            return self.deleteDuplicates(head)
+        
+        head.next = self.deleteDuplicates(head.next)
+        return head
+
+```
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