Leetcode123123. Best Time to Buy and Sell Stock III

此题限制交易次数为2次。

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).

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解法：思路是从2头分别求最大收益，最后遍历一次，求最大收益

class Solution {public:    int maxProfit(vector<int>& prices) {int n = prices.size();if (n <= 1)return 0;if (n == 2)return ((prices[1] - prices[0])>=0? prices[1]-prices[0]:0);int Max = prices[n - 1];int value = 0;vector<int> Maxprofit(n, 0);int i;for (i = n - 2; i >= 0; i--){if (prices[i]>Max)Max = prices[i];value = Max - prices[i];if (value>Maxprofit[i+1])Maxprofit[i] = value;elseMaxprofit[i] = Maxprofit[i + 1];}vector<int> Maxprofit2(n, 0);int Min = prices[0];for (i = 1; i <n; i++){if (prices[i]<Min)Min = prices[i];value = prices[i] - Min;if (value>Maxprofit2[i-1])Maxprofit2[i] = value;elseMaxprofit2[i] = Maxprofit2[i - 1];} int tmp = 0; int Max_result = 0;for (i = 0; i<n; i++){tmp = Maxprofit[i] + Maxprofit2[i];if (tmp>Max_result)        Max_result = tmp;}return  Max_result;}   };