111_leetcode_Best Time to Buy and Sell Stock III

最新推荐文章于 2025-09-10 22:03:14 发布

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本文探讨了在给定的股票价格数组中，如何通过最多进行两次买卖操作来实现最大利润的算法。利用两次遍历和动态规划方法，解决了一般情况下寻找最优交易策略的问题。

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).

1:特殊情况；2:从前到后和从后到前遍历两次数字；3:在遍历的过程中，保存以当前索引为结束点或者开始点的只进行一次买卖的最大收益；4:注意边界结束条件

        int maxProfit(vector<int> &prices)
    {
        if(prices.size() <= 1)
        {
            return 0;
        }
        
        int size = (int)prices.size();
        vector<int> leftProfit(size, 0);
        leftProfit[0] = 0;
        int minValue = prices[0];
        
        int result = 0;
        
        for(int i = 1; i < size; i++)
        {
            if(prices[i] > minValue)
            {
                leftProfit[i] = (prices[i] - minValue > leftProfit[i-1] ? prices[i] - minValue : leftProfit[i-1]);
            }
            else
            {
                minValue = prices[i];
                leftProfit[i] = leftProfit[i-1];
            }
        }
        
        result = leftProfit[size-1] > leftProfit[size-2] ? leftProfit[size-2] : leftProfit[size-1];
        int maxValue = prices[size -1];
        int rightMaxProfit = 0;
        
        for(int i = size - 2; i >= 0; i--)
        {
            if(prices[i] < maxValue)
            {
                rightMaxProfit = (rightMaxProfit > maxValue - prices[i] ? rightMaxProfit : maxValue - prices[i]);
            }
            else
            {
                maxValue = prices[i];
            }
            
            if(i == 0)
            {
                result = (result > rightMaxProfit ? result : rightMaxProfit);
            }
            else
            {
                result = (result > rightMaxProfit + leftProfit[i-1] ? result : rightMaxProfit + leftProfit[i-1]);
            }
        }
        
        return result;
    }