LEETCODE: Best Time to Buy and Sell Stock III

最新推荐文章于 2025-12-05 14:37:41 发布

原创最新推荐文章于 2025-12-05 14:37:41 发布 · 268 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#LEETCODE #遍历 #算法

LEETCODE 专栏收录该内容

153 篇文章

订阅专栏

本文探讨了如何通过最多进行两次股票交易来获取最大利润的算法。通过两次扫描，一次从左到右，另一次从右到左，算法能够计算出最优的买卖时机。实现方式涉及使用辅助向量存储每一步的最大差值，最终求和得到最大利润。

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.

有点难，不过和II的解法是一样的，不过要做两次扫描，从左往右，再从右往左。

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        int profit = 0;
        if (prices.size() == 0) {
            return 0;
        }
        vector<int> lefts(prices.size());
        vector<int> rights(prices.size());
        // Scan from left to right, to find current max gap from 0 to ii,
        // and store them in lefts.
        int min = prices[0];
        for (int ii = 1; ii < prices.size(); ii ++) {
            lefts[ii] = std::max(prices[ii] - min, lefts[ii - 1]);        
            min = std::min(prices[ii], min);
        }
        // Scan from right to left, to find current max gap from ii to size - 1,
        // and store them in rights.
        int max = prices[prices.size() - 1];
        for (int ii = prices.size() - 2; ii >= 0; ii --) {
            rights[ii] = std::max(max - prices[ii], rights[ii + 1]);
            max = std::max(prices[ii], max);
        }
        
        // Find the max sum of lefts and rights.
        for (int ii = 0; ii < prices.size(); ii++) {
            profit = std::max(lefts[ii] + rights[ii], profit);
        }
        return profit;      
    }
};