Lintcode393 Best Time to Buy and Sell Stock IV solution 题解

【题目描述】

Say you have an array for which the th element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete at most k transactions.

Notice:You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

假设你有一个数组,它的第i个元素是一支给定的股票在第i天的价格。

设计一个算法来找到最大的利润。你最多可以完成 k 笔交易。

【注】你不可以同时参与多笔交易(你必须在再次购买前出售掉之前的股票)

【题目链接】

www.lintcode.com/en/problem/best-time-to-buy-and-sell-stock-iv/

【题目解析】

下面的解法主要是能把两次的限制推广到k次交易:

这道题是Best Time to Buy and Sell Stock的扩展,现在最多可以进行两次交易。所以仍然使用动态规划来完成,事实上可以解决非常通用的情况,也就是最多进行k次交易的情况。 这里我们先解释最多可以进行k次交易的算法,然后最多进行两次我们只需要把k取成2即可。我们还是使用“局部最优和全局最优解法”。我们维护两种量,一个是当前到达第i天可以最多进行j次交易,最好的利润是多少(global[i][j]),另一个是当前到达第i天,最多可进行j次交易,并且最后一次交易在当天卖出的最好的利润是多少(local[i][j])。下面我们来看递推式,全局的比较简单,

global[i][j]=max(local[i][j],global[i-1][j]),

也就是去当前局部最好的,和过往全局最好的中大的那个(因为最后一次交易如果包含当前天一定在局部最好的里面,否则一定在过往全局最优的里面)。

全局(到达第i天进行j次交易的最大收益) = max{局部(在第i天交易后,恰好满足j次交易),全局(到达第i-1天时已经满足j次交易)}

对于局部变量的维护,递推式是

local[i][j]=max(global[i-1][j-1]+max(diff,0),local[i-1][j]+diff),

也就是看两个量,第一个是全局到i-1天进行j-1次交易,然后加上今天的交易,如果今天是赚钱的话(也就是前面只要j-1次交易,最后一次交易取当前天),第二个量则是取local第i-1天j次交易,然后加上今天的差值(这里因为local[i-1][j]比如包含第i-1天卖出的交易,所以现在变成第i天卖出,并不会增加交易次数,而且这里无论diff是不是大于0都一定要加上,因为否则就不满足local[i][j]必须在最后一天卖出的条件了)。

局部(在第i天交易后,总共交易了j次) = max{情况2,情况1}

情况1:在第i-1天时,恰好已经交易了j次(local[i-1][j]),那么如果i-1天到i天再交易一次:即在第i-1天买入,第i天卖出(diff),则这不并不会增加交易次数!【例如我在第一天买入,第二天卖出;然后第二天又买入,第三天再卖出的行为 和 第一天买入,第三天卖出 的效果是一样的,其实只进行了一次交易!因为有连续性】 情况2:第i-1天后,共交易了j-1次(global[i-1][j-1]),因此为了满足“第i天过后共进行了j次交易,且第i天必须进行交易”的条件:我们可以选择1:在第i-1天买入,然后再第i天卖出(diff),或者选择在第i天买入,然后同样在第i天卖出(收益为0)。

上面的算法中对于天数需要一次扫描,而每次要对交易次数进行递推式求解,所以时间复杂度是O(n*k),如果是最多进行两次交易,那么复杂度还是O(n)。空间上只需要维护当天数据皆可以,所以是O(k),当k=2,则是O(1)。

【参考答案】

www.jiuzhang.com/solutions/best-time-to-buy-and-sell-stock-iv/

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原文地址:https://www.cnblogs.com/rockerrrrrr/p/8949636.html

时间: 2024-11-08 11:05:56

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