leetcode_c++:Best Time to Buy and Sell Stock III(123)
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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 two transactions.
算法
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。最多交易两次,手上最多只能持有一支股票,求最大收益。
分析:动态规划法。以第i天为分界线,计算第i天之前进行一次交易的最大收益preProfit[i],和第i天之后进行一次交易的最大收益postProfit[i],最后遍历一遍,max{preProfit[i] + postProfit[i]} (0≤i≤n-1)就是最大收益。第i天之前和第i天之后进行一次的最大收益求法同Best Time to Buy and Sell Stock I。
代码:时间O(n),空间O(n)。
class Solution {public: int maxProfit(vector<int>& prices) { if(prices.size()<2) return 0; int n=prices.size(); vector<int> preprofile(n); vector<int> postprofile(n); int curmin=prices[0]; for(int i=1;i<n;i++){ curmin=min(curmin,prices[i]); preprofile[i]=max(preprofile[i-1],prices[i]-curmin); } int curmax=prices[n-1]; for(int i=n-2;i>=0;i--){ curmax=max(curmax,prices[i]); postprofile[i]=max(postprofile[i+1],curmax-prices[i]); } int maxprofile=0; for(int i=0;i<n;i++){ maxprofile=max(maxprofile,preprofile[i]+postprofile[i]); } return maxprofile; }};
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