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【leetcode】Best Time to Buy and Sell Stock III
Best Time to Buy and Sell Stock III
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).
设dpBack[i]为从第一天到第i天的最大的收益
设dpAfter[j]为从第j天到最后一天的最大收益
如何求最大收益可以考虑采用Best Time to Buy and Sell Stock I的方法
注意是最多买两次，也可以只买一次
1 class Solution { 2 public: 3 int maxProfit(vector<int> &prices) { 4 5 int n=prices.size(); 6 if(n==0)return 0; 7 8 vector<int> dpBack(n),dpAfter(n); 9 10 int maxProfit=0;11 dpBack[0]=0;12 int left=prices[0];13 14 for(int i=1;i<n;i++)15 {16 if(prices[i]>left)17 {18 if(maxProfit<prices[i]-left) maxProfit=prices[i]-left;19 }20 else21 {22 left=prices[i];23 }24 25 dpBack[i]=maxProfit;26 }27 28 maxProfit=0;29 dpAfter[n-1]=0;30 int right=prices[n-1];31 32 for(int j=n-2;j>=0;j--)33 {34 if(prices[j]<right)35 {36 if(maxProfit<right-prices[j]) maxProfit=right-prices[j];37 }38 else39 {40 right=prices[j];41 }42 43 dpAfter[j]=maxProfit;44 }45 46 int result=0;47 for(int i=0;i<n-1;i++)48 {49 if(dpBack[i]+dpAfter[i+1]>result) result=dpBack[i]+dpAfter[i+1];50 51 if(result<dpBack[i+1]) result=dpBack[i+1];52 }53 54 return result;55 }56 };
【leetcode】Best Time to Buy and Sell Stock III
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