leetcode 刷题之路 73 Best Time to Buy and Sell Stock III
2014-08-10 17:55
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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 two transactions.
Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
这次题目把购买次数限制为两次。根据题目规定,在第二次购买股票时现有的股票已经出售,若在第i天出售第一次购买的股票,那么可以把数组分为(0~i)和(i~n-1)两个独立的部分,可以分别使用两个数组maxPro,maxProRev来记录两部分的最大收益,maxPro[i]代表0~i天内最大的收益,maxProRev[i]代表i~n-1天内最大的收益,那么必然存在一个i,使得maxPro[i]+maxProRev[i]最大,这个最大值就是题目要求的结果。
我们在Best Time to Buy and Sell Stock I中求解过最大收益,只需要在这个基础上使用maxPro记录max值即可。
maxProRev[i]的求解可以通过从后往前的方式推导,由maxProRev[i+1]推导出maxProRev[i],具体逻辑如下:
使用max记录prices[i+1]~prices[n-1]中的最大值,也就是最高卖出价格
当max-prices[i]>maxProRev[i+1]时,更新最大值,否则,最大值还是maxProRev[i+1]
有了maxPro和maxProRev数组,我们只要遍历一次数组,求得的maxPro[i]+maxProRev[i]的最大值,就是所要的结果。
AC code:
Design an algorithm to find the maximum profit. You may complete at most two transactions.
Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
这次题目把购买次数限制为两次。根据题目规定,在第二次购买股票时现有的股票已经出售,若在第i天出售第一次购买的股票,那么可以把数组分为(0~i)和(i~n-1)两个独立的部分,可以分别使用两个数组maxPro,maxProRev来记录两部分的最大收益,maxPro[i]代表0~i天内最大的收益,maxProRev[i]代表i~n-1天内最大的收益,那么必然存在一个i,使得maxPro[i]+maxProRev[i]最大,这个最大值就是题目要求的结果。
我们在Best Time to Buy and Sell Stock I中求解过最大收益,只需要在这个基础上使用maxPro记录max值即可。
maxProRev[i]的求解可以通过从后往前的方式推导,由maxProRev[i+1]推导出maxProRev[i],具体逻辑如下:
使用max记录prices[i+1]~prices[n-1]中的最大值,也就是最高卖出价格
当max-prices[i]>maxProRev[i+1]时,更新最大值,否则,最大值还是maxProRev[i+1]
有了maxPro和maxProRev数组,我们只要遍历一次数组,求得的maxPro[i]+maxProRev[i]的最大值,就是所要的结果。
AC code:
class Solution { public: int maxProfit(vector<int> &prices) { if(prices.empty()) return 0; int n=prices.size(); int *maxPro=new int ; int *maxProRev=new int ; memset(maxPro,0,n*sizeof(int)); memset(maxProRev,0,n*sizeof(int)); int min=prices[0]; //最低买入价 int max=0; //最大利润 for(int i=0;i<n;i++) { if(min>prices[i]) min=prices[i]; if(max<prices[i]-min) max=prices[i]-min; maxPro[i]=max; //maxPro储存第i天的最大利润 } max=prices[n-1]; //出现过的最高卖出价 maxProRev[n-1]=0; for(int i=n-2;i>=0;i--) { if(max-prices[i]>maxProRev[i+1]) maxProRev[i]=max-prices[i]; else maxProRev[i]=maxProRev[i+1]; if(max<prices[i]) max=prices[i]; } max=0; for(int i=0;i<n;i++) { if(max<maxPro[i]+maxProRev[i]) max=maxPro[i]+maxProRev[i]; } delete []maxPro; delete []maxProRev; return max; } };
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