思路1

泛化到k次交易,秒杀所有股票交易问题

dp[k, i] = max(dp[k, i-1], prices[i] - prices[j] + dp[k-1, j-1]), j=[0..i-1]

dp[k, i] 表示k次交易,第i天卖(或不交易)时的最大收益

两种情况:

  • i天不交易,则dp[k,i] = dp[k, i-1]

  • i天交易(卖),假设我们在j天买了,则dp[k,i] = prices[i] - prices[j] + dp[k-1, j-1]

j是可以等于i的,j=i,则 dp[k,i] = max(dp[k,i-1], dp[k-1][i-1])

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