[LeetCode][Java] Best Time to Buy and Sell Stock IV
2017-05-13 10:31
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题目:
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:
给定一个数组,数组中第 i 个元素,表示给定的一仅仅股票在第 i 天的价格.设计一个算法找出最大的收益。
你最多被同意
k 次交易。
注意:
同一时间你不能进行多次交易。
(即:在你必须先卖掉这仅仅股票才干再次购买)
算法分析:
採用动态规划进行求解,使用局部最优和全局最优解法因为要考虑交易次数。维护量应该就是一个二维数组。
定义维护量:
global[i][j]:在到达第i天时最多可进行j次交易的最大利润。此为全局最优
local[i][j]:在到达第i天时最多可进行j次交易而且最后一次交易在最后一天卖出的最大利润,此为局部最优
定义递推式:
global[i][j]=max(global[i-1][j],local[i][j]);即第i天没有交易,和第i天有交易
local[i][j]=max(global[i-1][j-1]+max(diff,0),local[i-1][j]+diff)
diff=price[i]-price[i-1];
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。假设今天是赚钱的话(也就是前面仅仅要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(这里由于local[i-1][j]比方包括第i-1天卖出的交易,所以如今变成第i天卖出。并不会添加交易次数,
并且这里不管diff是不是大于0都一定要加上。由于否则就不满足local[i][j]必须在最后一天卖出的条件了)
这道题还有个坑,就是假设k的值远大于prices的天数。比方k是好几百万,而prices的天数就为若干天的话,上面的DP解法就很的没有效率,应该直接用
《Best
Time to Buy and Sell Stock II》的方法来求解。所以实际上这道题是之前的二和三的综合体。
AC代码:
<span style="font-family:Microsoft YaHei;font-size:12px;">public class Solution { public int maxProfit(int k, int[] prices) { if(prices==null || prices.length==0) return 0; if(k>prices.length)//k次数大于天数时,转化为问题《Best Time to Buy and Sell Stock II》--无限次交易的情景 { if(prices==null) return 0; int res=0; for(int i=0;i<prices.length-1;i++) { int degit=prices[i+1]-prices[i]; if(degit>0) res+=degit; } return res; } /* 定义维护量: global[i][j]:在到达第i天时最多可进行j次交易的最大利润,此为全局最优 local[i][j]:在到达第i天时最多可进行j次交易而且最后一次交易在最后一天卖出的最大利润,此为局部最优 定义递推式: global[i][j]=max(global[i-1][j],local[i][j]);即第i天没有交易,和第i天有交易 local[i][j]=max(global[i-1][j-1]+max(diff,0),local[i-1][j]+diff) diff=price[i]-price[i-1]; */ int[][] global=new int[prices.length][k+1]; int[][] local=new int[prices.length][k+1]; for(int i=0;i<prices.length-1;i++) { int diff=prices[i+1]-prices[i]; for(int j=0;j<=k-1;j++) { local[i+1][j+1]=Math.max(global[i][j]+Math.max(diff,0),local[i][j+1]+diff); global[i+1][j+1]=Math.max(global[i][j+1],local[i+1][j+1]); } } return global[prices.length-1][k]; } }</span>
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