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Rinta-Rinta
reviewed
Jan 11, 2026
Comment on lines
+79
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+91
| while (node != nullptr) { | ||
| if (node->next != nullptr && node->val == node->next->val) { | ||
| int dupVal = node->val; | ||
| while (node != nullptr && node->val == dupVal) { | ||
| node = node->next; | ||
| } | ||
| unique->next = node; | ||
| } else { | ||
| node = node->next; | ||
| unique = unique->next; | ||
| } | ||
| } | ||
| return dummy.next; |
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Rinta-Rinta/LeetCode_arai60#3 (comment)
https://discord.com/channels/1084280443945353267/1195700948786491403/1196701558382018590
処理が分岐する際に、処理の長さや簡単さによってはif-elseで並べずにcontinueやbreakで抜ける方が、
分かりやすい場合もありそうです。
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なるほど、大事なご指摘だと思います。
特にシンプルな処理の場合は、先にcontinueで抜けた方が可読性が上がる場面も多いと感じました。
ありがとうございます🙏
oda
reviewed
Jan 11, 2026
Comment on lines
+23
to
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| - headが重複している場合のエッジケースの考慮が必要 | ||
| - head一つのみのときの処理が必要 |
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この方法でもできるはずですが、確かに大変です。Step 2の方法で、重複していない先頭が見つかるまでは別のループを前につけるか、そうでなければ、保存するノードが見つかったときに、先頭が見つかっているかいなかで条件分岐します。
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なるほど、番兵を用意するか条件分岐で実現できますね。ありがとうございます。
愚直にも実装できるようにはしておきたいと思います。
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82. Remove Duplicates from Sorted List II