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35. Search Insert Position #39

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35. Search Insert Position #39

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11 changes: 11 additions & 0 deletions 0035-search-insert-position/memo.md
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### step1

findとlower_boundを使ってit != nums.end()のときはそのイテレータ、そうでない場合は配列のサイズを返すようにした。

### step2

他の方のコード（https://github.com/5ky7/arai60/pull/41/changes#diff-0c860cd754249868513e4f9054206317fa33d0f548fc3896ac2b3e11822fd852R19）を見てみると、lower_bound自体を実装していたので自分も実装してみた。
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@liquo-rice liquo-rice May 1, 2026

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実装したlower_boundのコードもPRに含めると良いと思います。

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@hemispherium hemispherium May 7, 2026

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すみません、何か勘違いしているかもなのですがstep2に載せたコードがlower_boundの実装のつもりでした。

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@liquo-rice liquo-rice May 7, 2026

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lower_boundと振る舞いが違うので、ドキュメントを確認してみると良さそうです。
https://en.cppreference.com/cpp/algorithm/lower_bound

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@hemispherium hemispherium May 8, 2026

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承知しました。ありがとうございます！


### step3

step2を3回通すまで書き直し。
14 changes: 14 additions & 0 deletions 0035-search-insert-position/step1.cpp
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class Solution {
public:
int searchInsert(vector<int>& nums, int target) {
auto it = find(nums.begin(), nums.end(), target);
if (it != nums.end()) {
return it - nums.begin();
}
Comment on lines +4 to +7
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@liquo-rice liquo-rice May 1, 2026

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問題文には、

You must write an algorithm with O(log n) runtime complexity.

とありますが、こちらの計算量はどうなっていますか？

lower_boundを一回呼び出すだけで書けますか？

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@hemispherium hemispherium May 7, 2026

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5ky7さんのPRを参考にlower_bound1回だけで書いてみました。見つからない場合もitがnums.end()になるのでdistanceを取ればちょうどnums.size()と同じになるのでこれだけでよかったです。
https://github.com/5ky7/arai60/pull/41/changes#diff-0c860cd754249868513e4f9054206317fa33d0f548fc3896ac2b3e11822fd852R5

class Solution {
public:
    int searchInsert(vector<int>& nums, int target) {
        auto it = lower_bound(nums.begin(), nums.end(), target);
        int position = distance(nums.begin(), it);
        return position;
    }
};

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@hemispherium hemispherium May 7, 2026

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ぼくの書いた方の計算量ですが、lower_boundの計算量はO(log n)ですがfindの方の計算量はO(n)のようなので問題文の条件をクリアできていなさそうです。

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@liquo-rice liquo-rice May 7, 2026

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見つからない場合もitがnums.end()になる

見つからない場合は、いつもそうなりますかね？

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@hemispherium hemispherium May 8, 2026

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見つからない場合というかtarget以上の要素がない場合、つまりtargetがいちばん後ろにくるべきケースではlower_boundでnums.end()が返ると思うのでこれでいいかなと思いました。

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@liquo-rice liquo-rice May 8, 2026

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「（target以上の要素が）見つからない」という意味でしたら、それで合っていますね。

auto it2 = lower_bound(nums.begin(), nums.end(), target);
if (it2 != nums.end()) {
return it2 - nums.begin();
}
return nums.size();
}
};
20 changes: 20 additions & 0 deletions 0035-search-insert-position/step2.cpp
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class Solution {
public:
int searchInsert(vector<int>& nums, int target) {
int left = 0;
int right = nums.size();

while (left < right) {
int mid = (left + right) / 2;
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@nodchip nodchip May 1, 2026

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オーバーフローを避けるため、

int mid = left + (right - left) / 2;

と書いたほうが良いと思います。

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@hemispherium hemispherium May 7, 2026

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承知しました。確かにそちらの方がよさそうです。

if (nums[mid] == target) {
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@nodchip nodchip May 1, 2026

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念のため確認させてください。この if 文はなくても動きますか？

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@hemispherium hemispherium May 7, 2026

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あまり深く考えていなかったのですが、二分探索が終わるとleftとrightの値が同じになるはずなのでこのif文は必要なかったと思います。一応LeetCodeでも確認したところなくても動きました。

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@liquo-rice liquo-rice May 7, 2026

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二分探索が終わるとleftとrightの値が同じになる

とありますが、ループ終了後には、どこを指していますか？

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@hemispherium hemispherium May 8, 2026

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以下のプレイグラウンドで値を見ながら確認したのですが、nums[mid]がtarget以上の場合はright = midに、nums[mid]がtargetより小さい場合はleft = mid + 1になるので最終的にleftとrightは同じ値になって、target以上の最小のnumのインデックスを指しそうです。
https://leetcode.com/playground/mR6MWyoh

また、下記のコードがあることで最終的な答えは変わらないですがnums[mid] == targetの場合は早期にループを抜けられることもわかりました。

if (nums[mid] == target) {
    return mid;
}

return mid;
}
if (nums[mid] < target) {
left = mid + 1;
} else {
right = mid;
}
}
return left;
}
};
20 changes: 20 additions & 0 deletions 0035-search-insert-position/step3.cpp
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class Solution {
public:
int searchInsert(vector<int>& nums, int target) {
int left = 0;
int right = nums.size();

while (left < right) {
int mid = (left + right) / 2;
if (nums[mid] == target) {
return mid;
}
if (nums[mid] < target) {
left = mid + 1;
} else {
right = mid;
}
}
return left;
}
};