198. House Robber#33
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nodchip
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Apr 23, 2026
| class Solution { | ||
| public: | ||
| int rob(vector<int>& nums) { | ||
| vector<int> sum(nums.size()); |
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2 つ前の値までしか参照していないため、変数を 2 つ使ってループのイテレーション間で値を伝搬することで、追加の空研計算量を O(1) とすることがあります。もし余裕があれば試してみるとよいと思います。
| int rob(vector<int>& nums) { | ||
| vector<int> sum(nums.size()); | ||
| for (int i = 0; i < nums.size(); i++) { | ||
| if (i == 0) { |
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i == 0 のケースと i == 1 のケースをループの外で処理することで、ループの内容をシンプルにすることができると思います。
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そうですね、i==0とi==1のケースはfor文の外に出した方がよさそうです。
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この問題：https://leetcode.com/problems/house-robber/description/
次に解く問題：https://leetcode.com/problems/house-robber-ii/description/