Best Time to Buy and Sell Stock III

最后更新于:2022-04-01 22:57:20

**一. 题目描述** 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). **二. 题目分析** 和前两道题相比,这道题限制了股票的交易次数,最多只能交易两次。 可使用动态规划来完成,首先是进行第一步扫描,先计算出序列`[0, …, i]`中的最大利润`profit`,用一个数组`f1`保存下来,这一步时间复杂度为`O(n)`。 第二步是逆向扫描,计算子序列`[i, …, n - 1]`中的最大利润`profit`,同样用一个数组`f2`保存下来,这一步的时间复杂度也是`O(n)`。 最后一步,对于,对`f1 + f2`,找出最大值即可。 **三. 示例代码** ~~~ #include #include using namespace std; class Solution { public: int maxProfit(vector &prices) { int size = prices.size(); if (size <= 1) return 0; vector f1(size); vector f2(size); int minV = prices[0]; for (int i = 1; i < size; ++i) { minV = std::min(minV, prices[i]); f1[i] = std::max(f1[i - 1], prices[i] - minV); } int maxV = prices[size - 1]; f2[size - 1] = 0; for (int i = size-2; i >= 0; --i) { maxV = std::max(maxV, prices[i]); f2[i] = std::max(f2[i + 1], maxV - prices[i]); } int sum = 0; for (int i = 0; i < size; ++i) sum = std::max(sum, f1[i] + f2[i]); return sum; } }; ~~~ **四. 小结** 相比前两题,该题难度稍大,与该题相关的题目有好几道。后续更新…
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