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70. Climbing Stairs #16
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,19 @@ | ||
| # 70. Climbing Stairs | ||
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| https://leetcode.com/problems/climbing-stairs/ | ||
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| ## Comments | ||
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| ### step1 | ||
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| * dp や再帰で解けるだろう | ||
| * いくつか試すと、直近 2 つだけ持っていればよい | ||
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| ### step2 | ||
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| * フィボナッチっぽく実装するならこのあたりとか | ||
| * for のループ回数とか境界条件をちゃんと確認する必要はある | ||
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| ### step3 | ||
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| * 簡単な問題なので省略。ただし LeetCode に色々別解はある。 | ||
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,20 @@ | ||
| class Solution { | ||
| public: | ||
| int climbStairs(int n) { | ||
| if (n <= 1) { | ||
| return n; | ||
| } | ||
| int ways; | ||
| // Starting from ways(0) = 0 | ||
| int two_steps_before = 0; | ||
| // ways(1) = 1 | ||
| int one_step_before = 1; | ||
| // Run n times. | ||
| for (int i = 0; i < n; ++i) { | ||
| ways = two_steps_before + one_step_before; | ||
| two_steps_before = one_step_before; | ||
| one_step_before = ways; | ||
| } | ||
| return ways; | ||
| } | ||
| }; |
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,16 @@ | ||
| class Solution { | ||
| public: | ||
| int climbStairs(int n) { | ||
| if (n <= 1) { | ||
| return n; | ||
| } | ||
| int first = 1; | ||
| int second = 1; | ||
| for (int i = 2; i <= n; ++i) { | ||
| int third = first + second; | ||
| first = second; | ||
| second = third; | ||
| } | ||
| return second; | ||
| } | ||
| }; |
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ぱっと思いつくのは、
あたりでした。自分が面接官なら、動的計画法までを期待値とすると思います。すらすら解けたら、行列の n 乗の方法を説明したうえで、 m * m 正方行列の乗算のコードを書けるかどうか試すと思います。