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Merged

pull merged 5 commits into AlgorithmAndLeetCode:master from labuladong:master

Jun 9, 2023

Merged

[pull] master from labuladong:master #98

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fix code: java 22

kkty39 Jun 4, 2023

fix code: java 22

kkty39 Jun 4, 2023

fix typo in comment

kkty39 Jun 4, 2023

remove fixes in another problem

kkty39 Jun 4, 2023

labuladong Jun 9, 2023

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2 changes: 1 addition & 1 deletion 动态规划系列/团灭股票问题.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -140,7 +140,7 @@ dp[i][k][1] = max(dp[i-1][k][1], dp[i-1][k-1][0] - prices[i])

		2、我昨天本没有持有,且截至昨天最大交易次数限制为 `k - 1`;但今天我选择 `buy`,所以今天我就持有股票了,最大交易次数限制为 `k`。

	> note:这里着重提醒一下,时刻牢记「状态」的定义,状态 `k` 的定义并不是「已进行的交易次数」,而是「最大交易次数的上限限制」。如果确定今天进行一次交易,且要保证截至今天最大交易次数上限为 `k`,那么昨天的最大交易次数上限必须是 `k - 1`。
	> note:这里着重提醒一下,时刻牢记「状态」的定义,状态 `k` 的定义并不是「已进行的交易次数」,而是「最大交易次数的上限限制」。如果确定今天进行一次交易,且要保证截至今天最大交易次数上限为 `k`,那么昨天的最大交易次数上限必须是 `k - 1`。举个具体的例子,比方说要求你的银行卡里今天至少有 100 块钱,且你确定你今天可以赚 10 块钱,那么你就要保证昨天的银行卡要至少剩下 90 块钱。

		这个解释应该很清楚了,如果 `buy`,就要从利润中减去 `prices[i]`,如果 `sell`,就要给利润增加 `prices[i]`。今天的最大利润就是这两种可能选择中较大的那个。

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