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hayashi-ay
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Mar 11, 2024
| int currentOther, index; | ||
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| for(index = 0; index < nums.size(); index++){ | ||
| currentOther = target - nums[index]; |
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もっといい命名がある気がします。「現在の他」とはどういうことだろうと思いました。pair、complement、numToFindあたりでしょうか？もっと良いのがあるかもしれないですが。
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回答となる組を作るイメージで、nums[i]をcurrentでその片方という意味でcurrentOtherとしてましたが
numToFindのほうがわかりやすいですね
| for(index = 0; index < nums.size(); index++){ | ||
| currentOther = target - nums[index]; | ||
| if(numMap.find(currentOther) != numMap.end()){ | ||
| break; |
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（小田さんのコメントと重複しますが）ここでreturnしてあげれば、int currentOther, indexのスコープをfor文に限定できます。これくらいのコードであれば、変数の数も少ないのでスコープを限定しなくても書けちゃいますが、ある程度複雑になると不必要に脳内のワーキングメモリを使うことになります。
| numMap[nums[index]] = index; | ||
| } | ||
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| return { index, numMap[currentOther] }; |
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（これも小田さんのコメントと重複しますが）課題制約で解があることは保証されていますが、解がない場合も考慮してみるとどうでしょうか。
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https://leetcode.com/problems/two-sum/