Best Time to Buy and Sell Stock IV
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Description: Say you have an array for which the ith 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. Note: You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
题意:用一个数组表示股票每天的价格,数组的第i个数表示股票在第i天的价格。最多交易k次,手上最多只能持有一支股票,求最大收益。
分析:特殊动态规划法。传统的动态规划我们会这样想,到第i天时进行j次交易的最大收益,要么等于到第i-1天时进行j次交易的最大收益(第i天价格低于第i-1天的价格),要么等于到第i-1天时进行j-1次交易,然后第i天进行一次交易(第i天价格高于第i-1天价格时)。于是得到动规方程如下(其中diff = prices[i] – prices[i – 1]):
profit[i][j] = max(profit[i – 1][j], profit[i – 1][j – 1] + diff)
看起来很有道理,但其实不对,为什么不对呢?因为diff是第i天和第i-1天的差额收益,如果第i-1天当天本身也有交易呢(也就是说第i-1天刚卖出了股票,然后又买入等到第i天再卖出),那么这两次交易就可以合为一次交易,这样profit[i – 1][j – 1] + diff实际上只进行了j-1次交易,而不是最多可以的j次,这样得到的最大收益就小了。
那么怎样计算第i天进行交易的情况的最大收益,才会避免少计算一次交易呢?我们用一个局部最优解和全局最有解表示到第i天进行j次的收益,这就是该动态规划的特殊之处。
用local[i][j]表示到达第i天时,最多进行j次交易的局部最优解;用global[i][j]表示到达第i天时,最多进行j次的全局最优解。它们二者的关系如下(其中diff = prices[i] – prices[i – 1]):
local[i][j] = max(global[i – 1][j – 1] , local[i – 1][j] + diff)
global[i][j] = max(global[i – 1][j], local[i][j])
local[i][j]和global[i][j]的区别是:local[i][j]意味着在第i天一定有交易(卖出)发生,当第i天的价格高于第i-1天(即diff > 0)时,那么可以把这次交易(第i-1天买入第i天卖出)跟第i-1天的交易(卖出)合并为一次交易,即local[i][j]=local[i-1][j]+diff;当第i天的价格不高于第i-1天(即diff<=0)时,那么local[i][j]=global[i-1][j-1]+diff,而由于diff<=0,所以可写成local[i][j]=global[i-1][j-1]。global[i][j]就是我们所求的前i天最多进行k次交易的最大收益,可分为两种情况:如果第i天没有交易(卖出),那么global[i][j]=global[i-1][j];如果第i天有交易(卖出),那么global[i][j]=local[i][j]。
参考:http://liangjiabin.com/blog/2015/04/leetcode-best-time-to-buy-and-sell-stock.html
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