[LeetCode] 010: Best Time to Buy and Sell Stock III

最新推荐文章于 2025-04-18 22:24:21 发布

1.02^365的成长裂变

最新推荐文章于 2025-04-18 22:24:21 发布

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本文介绍了一种算法，用于计算给定股票价格数组中通过至多两笔交易获得的最大利润。该算法首先计算从左到右的最佳卖出时机，再逆向寻找从右到左的最佳买入时机，最终确定最大收益。

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[Problem]

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).

[Solution]

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        // empty array
        if(prices.size() == 0){
            return 0;
        }
        
        // variables definition
        int left[prices.size()];
        int right[prices.size()];
        int minPrice = prices[0], maxPrice = prices[prices.size()-1];
        
        // update left
        int profit = 0;
        for(int i = 0; i < prices.size(); ++i){
            left[i] = prices[i] - minPrice;
            if(left[i] > profit){
                profit = left[i];
            }
            else{
                left[i] = profit;
            }
            minPrice = min(minPrice, prices[i]);
        }
        
        // update right
        profit = 0;
        for(int i = prices.size()-1; i >= 0; --i){
            right[i] = maxPrice - prices[i];
            if(right[i] > profit){
                profit = right[i];
            }
            else{
                right[i] = profit;
            }
            maxPrice = max(maxPrice, prices[i]);
        }
        
        // get the max profit
        int res = 0;
        for(int i = 0; i <= prices.size(); ++i){
            profit = 0;
            if(i-1 >= 0 && i-1 < prices.size()){
                profit += left[i-1];
            }
            if(i < prices.size()){
                profit += right[i];
            }
            res = max(res, profit);
        }
        return res;
    }
};