From e7755dbd4e18d70e850615ceaaa194b6545ad27e Mon Sep 17 00:00:00 2001
From: soma <140805585+Exzrgs@users.noreply.github.com>
Date: Sat, 19 Oct 2024 23:17:11 +0900
Subject: [PATCH] Revert "Add Two Numbers"

---
 arai60/add-two-numbers/step1.py | 33 --------------------
 arai60/add-two-numbers/step2.py | 53 ---------------------------------
 arai60/add-two-numbers/step3.py | 27 -----------------
 arai60/add-two-numbers/step4.py | 24 ---------------
 4 files changed, 137 deletions(-)
 delete mode 100644 arai60/add-two-numbers/step1.py
 delete mode 100644 arai60/add-two-numbers/step2.py
 delete mode 100644 arai60/add-two-numbers/step3.py
 delete mode 100644 arai60/add-two-numbers/step4.py

diff --git a/arai60/add-two-numbers/step1.py b/arai60/add-two-numbers/step1.py
deleted file mode 100644
index b643a96..0000000
--- a/arai60/add-two-numbers/step1.py
+++ /dev/null
@@ -1,33 +0,0 @@
-'''
-nextを作る位置を工夫しないと最後に余計なノードを作ってしまう
-最後にc==1で繰り上がりを足さないといけないことに最初は気づいていたが、実装しているうちに忘れてしまっていた...
-'''
-
-class Solution:
-    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
-        head = ListNode(-1)
-        now = head
-        
-        c = 0
-        while l1 and l2:
-            s = (l1.val + l2.val + c)%10
-            c = (l1.val + l2.val + c)//10
-            now.next = ListNode(s)
-            now = now.next
-            l1 = l1.next
-            l2 = l2.next
-        while l1:
-            s = (l1.val + c)%10
-            c = (l1.val + c)//10
-            now.next = ListNode(s)
-            now = now.next
-            l1 = l1.next
-        while l2:
-            s = (l2.val + c)%10
-            c = (l2.val + c)//10
-            now.next = ListNode(s)
-            now = now.next
-            l2 = l2.next
-        if c == 1:
-            now.next = ListNode(1)
-        return head.next
diff --git a/arai60/add-two-numbers/step2.py b/arai60/add-two-numbers/step2.py
deleted file mode 100644
index 4cda706..0000000
--- a/arai60/add-two-numbers/step2.py
+++ /dev/null
@@ -1,53 +0,0 @@
-'''
-headという変数名は答えのheadを表していそうだから、fake_headに変える
-nowもtailに変える
-whileが多いからまとめる。
-    ただ、実際1つのループにまとめるのが困難なケースもある気がするので、そこは実装する前に考えたほうが良いかもしれない
-
-口頭説明シミュレーション
-    まず、fake_headを作ります。
-    あとから後ろからノードをたどっていくのは大変なので、素早くheadを指せるポインタが欲しいからです。
-    次に、現在処理しているノードを表すtailを用意します。
-    また、l1とl2を足したときに繰り上がりが発生する可能性があるので、carryもここで用意しておきます。
-    l1かl2がNoneでないか、carryが1の間ループをします。
-    carryを入れる理由は、最後に繰り上がりだけ残った場合をループの中で処理したいからです。
-    そしてsum_valという変数は、それらを足した値とします。
-    carryを桁あふれ、つまりsum_valを10で割った商とします。
-    また、sum_valの10で割ったあまりをtailの次のノードの値とします。
-    そして、tail, l1, l2を次のステップへと進めます。
-    こうすることで、問題を解くことができました。
-    最後に桁上りしてl1とl2が終了する場合や、空のL1,l2を渡された場合にも対処できていることも確認できます。
-    
-    Firstly, we create "fake_head".
-    This is because we want a pointer that can quickly point to the head.
-    Next, we prepare "tail" that represents the node currently being processed.
-    We also need to prepare "carry" because there is possibility of carry-over when l1 and l2 added.
-    We loop while l1 or l2 is not None or if carry is 1.
-    The reason for including "carry" is that we want to handle the case where only the carry remains in the loop.
-    The variable "sum_val" is the value obtained by adding them together.
-    Let carry be the overflow. In other words, it is the quotient of sum_val divided by 10.
-    And let sum_val's remainder divided by 10 be the value of the next node of tail.
-    Then we proceed to the next step for tail, l1, and l2.
-    In this way, we have solved the problem.
-    We can also confirm that we are able to deal with the case where l1 and l2 terminate after overflow or when an empty l1 and l2 are passed.
-'''
-
-class Solution:
-    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
-        fake_head = ListNode(-1)
-        tail = fake_head
-
-        carry = 0
-        while l1 or l2 or carry:
-            sum_val = carry
-            if l1:
-                sum_val += l1.val
-                l1 = l1.next
-            if l2:
-                sum_val += l2.val
-                l2 = l2.next
-            carry = sum_val//10
-            tail.next = ListNode(sum_val%10)
-            tail = tail.next
-
-        return fake_head.next
diff --git a/arai60/add-two-numbers/step3.py b/arai60/add-two-numbers/step3.py
deleted file mode 100644
index c671a47..0000000
--- a/arai60/add-two-numbers/step3.py
+++ /dev/null
@@ -1,27 +0,0 @@
-'''
-意識したこと
-・fake_headを作る必要性
-確認したこと
-・各ポインタをちゃんと次のステップへ移動させているか?
-・l1やl2が空のときにしっかりと動くか?
-'''
-
-class Solution:
-    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
-        fake_head = ListNode(-1)
-        tail = fake_head
-
-        carry = 0
-        while l1 or l2 or carry:
-            sum_val = carry
-            if l1:
-                sum_val += l1.val
-                l1 = l1.next
-            if l2:
-                sum_val += l2.val
-                l2 = l2.next
-            carry = sum_val//10
-            tail.next = ListNode(sum_val%10)
-            tail = tail.next
-
-        return fake_head.next
diff --git a/arai60/add-two-numbers/step4.py b/arai60/add-two-numbers/step4.py
deleted file mode 100644
index f41ed86..0000000
--- a/arai60/add-two-numbers/step4.py
+++ /dev/null
@@ -1,24 +0,0 @@
-'''
-変数名をfake_head→dummy
-%の両端にspace
-'''
-
-class Solution:
-    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
-        dummy = ListNode(-1)
-        tail = dummy
-
-        carry = 0
-        while l1 or l2 or carry:
-            sum_val = carry
-            if l1:
-                sum_val += l1.val
-                l1 = l1.next
-            if l2:
-                sum_val += l2.val
-                l2 = l2.next
-            carry = sum_val // 10
-            tail.next = ListNode(sum_val % 10)
-            tail = tail.next
-
-        return dummy.next