[LeetCode]Best Time to Buy and Sell Stock III-优快云博客

本文介绍了一种算法，用于确定在给定股价数组的情况下，完成至多两次买卖交易所能获得的最大利润。通过预先计算买入和卖出的最佳时机，并利用动态规划的思想减少重复计算，实现了时间复杂度为O(n)的解决方案。

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Say you have an array for which the i^th 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).

Have you been asked this question in an interview? Yes

思考：最多交易两次，设断点分别求前后profit，相加返回最大值，O(n²)超时。原因是重复计算，参考这里，空间换时间。哎，还是思维太局限了，不知变通。

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        int len=prices.size();
        if(len==0||len==1) return 0;
        vector<int> profit1(len);
        vector<int> profit2(len);
        int i;
        profit1[0]=0;
        int buy=prices[0];
        for(i=1;i<len;i++)
        {
            buy=min(buy,prices[i]);
            profit1[i]=max(profit1[i-1],prices[i]-buy);
        }
        profit2[len-1]=0;
        int sell=prices[len-1];
        for(i=len-2;i>=0;i--)
        {
            sell=max(sell,prices[i]);
            profit2[i]=max(profit2[i+1],sell-prices[i]);
        }
        int profit=0;
        for(i=0;i<len;i++)
        {
            if(profit<(profit1[i]+profit2[i]))
            profit=profit1[i]+profit2[i];
        }
        return profit;
    }
};

转载于:https://www.cnblogs.com/Rosanna/p/3591516.html