diff --git a/46_permutations.md b/46_permutations.md new file mode 100644 index 0000000..2042271 --- /dev/null +++ b/46_permutations.md @@ -0,0 +1,240 @@ +# 46. Permutations + +## 1st + +### ① + +[cpythonのpermutationの内容](https://github.com/python/cpython/blob/bc264eac3ad14dab748e33b3d714c2674872791f/Modules/itertoolsmodule.c#L2603)を思い出しながら書いた。ギリギリ何とか書けたという感じ。 + +所要時間: 23:52 + +n: len(nums) +- 時間計算量: O(n*n!) +- 空間計算量: O(n!) (出力でn!個返すため) + +```py +class Solution: + def permute(self, nums: List[int]) -> List[List[int]]: + return list(self._perms(nums)) + + def _perms(self, nums: list[int]) -> Iterator[list[int]]: + length = len(nums) + indices = list(range(length)) + yield [nums[i] for i in indices] + cycles = list(range(length, 0, -1)) + while length: + for i in reversed(range(length)): + cycles[i] -= 1 + if cycles[i] == 0: + indices[i:] = indices[i+1:] + indices[i:i+1] + cycles[i] = length - i + else: + j = cycles[i] + indices[i], indices[-j] = indices[-j], indices[i] + yield [nums[i] for i in indices] + break + else: + return +``` + +### ② + +上の解法はnPrの各要素を計算をするための方法。r==nでいいならもっと簡単に書ける。 +`if i in indices` の部分は、どうせnが小さいので辞書にしてO(1)にしてもオーバーヘッド分で相殺されるかなと思い最適化はしていない。 + +`if len(indices) == len(nums)` の終わりはreturnすればよかった。 + +所要時間: 8:11 + +n: len(nums) +- 時間計算量: O(n^2 * n!) +- 空間計算量: O(n!) + +```py +class Solution: + def permute(self, nums: List[int]) -> List[List[int]]: + def perm(indices: list[int]) -> Iterator[list[int]]: + if len(indices) == len(nums): + yield [nums[i] for i in indices] + for i in range(len(nums)): + if i in indices: + continue + indices.append(i) + yield from perm(indices) + indices.pop() + + return list(perm([])) +``` + +### ③ + +1. 後ろから初めて昇順になっているindexを探す +2. 後ろから、1で見つけたindexの要素より初めて大きくなる要素のindexを探す +3. 上で見つけた2つのindexの位置を交換する +4. 1のindexより右側を逆順にする (アルゴリズム上、必ず降順になっているので逆にしたら辞書順最小になる) + +所要時間: 17:48 + +n: len(nums) +- 時間計算量: O(n*n!) +- 空間計算量: O(n!) + +```py +class Solution: + def permute(self, nums: List[int]) -> List[List[int]]: + if len(nums) <= 1: + return [nums[:]] + + def perm(indices: list[int]) -> Iterator[list[int]]: + i = len(nums) - 2 + while i >= 0 and indices[i] > indices[i + 1]: + i -= 1 + if i == -1: + return + j = len(indices) - 1 + while indices[i] > indices[j]: + j -= 1 + indices[i], indices[j] = indices[j], indices[i] + indices[i+1:] = reversed(indices[i+1:]) + yield [nums[i] for i in indices] + yield from perm(indices) + + return [nums[:]] + list(perm(list(range(len(nums))))) +``` + +## 2nd + +### 参考 + +- https://discord.com/channels/1084280443945353267/1226508154833993788/1255879916113756202 + - https://github.com/nittoco/leetcode/pull/21 + + +[e_1, e_2, ..., e_k] のリストがあるとき、要素の順番を保存したままe_{k+1}を挿入すると +- [e_1, e_2, ..., e_k, e_{k+1}] +- [e_1, e_2, ..., e_{k+1}, e_k] +- ... +- [e_{k+1}, e_1, e_2, ..., e_k] + +のk+1通りのリストができる。これをすべての要素の順番について列挙すれば、[e_1, ..., e_{k+1}]の要素を使ったすべての順列を書き出すことができる。 + +```py +class Solution: + def permute(self, nums: List[int]) -> List[List[int]]: + def perm(index: int) -> Iterator[list[int]]: + if index == len(nums): + yield [] + return + for permutation in perm(index + 1): + permutation.append(nums[index]) + yield permutation[:] + for i in reversed(range(len(permutation) - 1)): + permutation[i], permutation[i+1] = permutation[i+1], permutation[i] + yield permutation[:] + + return list(perm(0)) +``` + +ループでall_permutationsを更新していくやり方。変数名に自信はない。 + +```py +class Solution: + def permute(self, nums: List[int]) -> List[List[int]]: + all_permutations = [[]] + for num in reversed(nums): + derived_permutations = [] + for permutation in all_permutations: + permutation.append(num) + copied = permutation[:] + for i in reversed(range(len(copied) - 1)): + copied[i], copied[i+1] = copied[i+1], copied[i] + derived_permutations.append(copied[:]) + all_permutations += derived_permutations + return all_permutations +``` + +- https://discord.com/channels/1084280443945353267/1233295449985650688/1245243664259878932 + - https://github.com/Exzrgs/LeetCode/pull/19 +- https://discord.com/channels/1084280443945353267/1196472827457589338/1241302279303204915 + - https://github.com/Mike0121/LeetCode/pull/14 + + +順列の置換を列挙すれば良い。個人的には一番分かりやすい気がする。 + +スライスによるindicesのコピーはしておくとスレッドセーフになる (はず)。マルチスレッドにするならコードも変えないといけないが... + +```py +class Solution: + def permute(self, nums: List[int]) -> List[List[int]]: + if not nums: + return [[]] + + def perm(indices: list[int], i: int) -> Iterator[list[int]]: + if i == len(nums) - 1: + yield [nums[j] for j in indices] + return + for j in range(i, len(nums)): + indices[i], indices[j] = indices[j], indices[i] + yield from perm(indices[:], i + 1) + indices[i], indices[j] = indices[j], indices[i] + + indices = list(range(len(nums))) + return list(perm(indices[:], 0)) +``` + +- https://discord.com/channels/1084280443945353267/1233603535862628432/1237775088762490900 + - https://github.com/goto-untrapped/Arai60/pull/16 + + +スタックでも書いてみる。 + +```py +class Solution: + def permute(self, nums: List[int]) -> List[List[int]]: + if not nums: + return [[]] + stack = [(list(range(len(nums))), 0)] + all_permutations = [] + while stack: + indices, index = stack.pop() + if index == len(nums) - 1: + all_permutations.append([nums[i] for i in indices]) + continue + for i in range(index, len(nums)): + indices[index], indices[i] = indices[i], indices[index] + stack.append((indices[:], index + 1)) + indices[index], indices[i] = indices[i], indices[index] + return all_permutations +``` + +- https://discord.com/channels/1084280443945353267/1225849404037009609/1235272786474303560 + - https://github.com/SuperHotDogCat/coding-interview/pull/12 +- https://discord.com/channels/1084280443945353267/1201211204547383386/1228915333310713856 + - https://github.com/shining-ai/leetcode/pull/50 +- https://discord.com/channels/1084280443945353267/1200089668901937312/1220995877624217620 + - https://github.com/hayashi-ay/leetcode/pull/57 + + +## 3rd + +全体的に命名をサボりがちかも。ただこのくらいなら読めるんじゃないか？という気もしてこうなっている (が、後で読んだら読めないとかあるのかもしれない) + + +```py +class Solution: + def permute(self, nums: List[int]) -> List[List[int]]: + if not nums: + return [[]] + + def perm(indices: list[int], i: int) -> Iterator[list[int]]: + if i == len(nums) - 1: + yield [nums[j] for j in indices] + return + for j in range(i, len(nums)): + indices[i], indices[j] = indices[j], indices[i] + yield from perm(indices[:], i + 1) + indices[i], indices[j] = indices[j], indices[i] + + indices = list(range(len(nums))) + return list(perm(indices[:], 0)) +```