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Merged

pull merged 1 commit into AlgorithmAndLeetCode:main from itcharge:main

Jul 22, 2022

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[pull] main from itcharge:main #17

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Update 01.Enumeration-Algorithm.md

itcharge Jul 22, 2022

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2 changes: 1 addition & 1 deletion Contents/09.Algorithm-Base/01.Enumeration-Algorithm/01.Enumeration-Algorithm.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -186,7 +186,7 @@ class Solution:
		\| 二进制数对应位数 \| 0 \| 1 \| 0 \| 0 \| 1 \|
		\| 对应选取状态 \| 未选取 \| 选取 \| 未选取 \| 未选取 \| 选取 \|

	通过上面的例子我们可以得到启发:对于长度为 `5` 的集合 `S` 来说,我们只需要从 `00000` ~ `11111` 枚举一次(对应十进制为 0ドル \sim 2^4 - 1$)即可得到长度为 `5` 的集合 `S` 的所有子集。
	通过上面的例子我们可以得到启发:对于长度为 `5` 的集合 `S` 来说,我们只需要从 `00000` ~ `11111` 枚举一次(对应十进制为 0ドル \sim 2^5 - 1$)即可得到长度为 `5` 的集合 `S` 的所有子集。

		我们将上面的例子拓展到长度为 `n` 的集合 `S`。可以总结为:

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