回答編集履歴
2
追記
answer
CHANGED
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@@ -15,12 +15,41 @@
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15
15
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current = head;
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16
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while(current != tail){
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17
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if(current->birthday % 2 == 0){
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18
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-
Node* del = current;
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*now = current->next;
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20
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-
delete
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+
delete current;
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21
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} else {
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now = ¤t->next;
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}
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23
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current = *now;
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}
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26
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-
```
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25
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+
```
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26
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+
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27
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+
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28
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+
---
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29
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+
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30
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+
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31
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なんとなく、一応解説
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32
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+
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33
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```c
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34
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Node keeper;
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35
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keeper.next = head;
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36
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Node* last = &keeper;
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37
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// last は最後に調べた要素
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38
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while(last->next != tail){
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39
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current = last->next;
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40
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if(current->birthday % 2 == 0){
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41
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// currentの誕生日は偶数なので飛ばす
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42
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last->next = current->next;
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43
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delete current;
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44
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// last->nextを調べてないのでlastを更新しない
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45
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}
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46
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else
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47
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// last->nextを調べたのでlastを更新する
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48
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last = current;
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49
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}
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50
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head = keeper.next;
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51
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```
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52
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+
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53
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+
片方向リストは先頭要素を指すダミーを作ってしまうと↑のようになります。
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54
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+
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55
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よくみると、`last`は`last->next`にしか使ってないのでポインタでどうにかなるってのが提示したソースです。
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1
コード修正
answer
CHANGED
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@@ -14,13 +14,13 @@
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14
14
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Node** now = &head;
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15
15
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current = head;
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16
16
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while(current != tail){
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17
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-
if(current->birthday % 2 == 0){
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17
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+
if(current->birthday % 2 == 0){
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18
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-
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18
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+
Node* del = current;
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19
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-
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19
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*now = current->next;
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20
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-
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20
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delete del;
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21
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-
} else {
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21
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} else {
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22
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-
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22
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now = ¤t->next;
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23
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-
}
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23
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+
}
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24
24
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current = *now;
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25
25
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}
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26
26
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```
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