LeetCode 123: Best Time to Buy and Sell Stock III

Best Time to Buy and Sell Stock III

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).

解题思路

思路一：动态规划法。

以第i天为分界线，计算第 i 天之前进行一次交易的最大收益 preProfit[i]，和第 i 天之后进行一次交易的最大收益 postProfit[i]。最后遍历一遍，max(preProfit[i]+postProfit[i]),(0≤i≤n−1) 就是最大收益。第 i 天之前和第 i 天之后进行一次的最大收益求法同 LeetCode 121。

代码如下：

class Solution {public:    int maxProfit(vector<int>& prices) {        if (prices.size() < 2) return 0;        int n = prices.size();        int preProfit[n], postProfit[n];        preProfit[0] = 0;        int curMin = prices[0];        for (int i = 1; i < n; ++i) {            preProfit[i] = max(preProfit[i-1], prices[i] - curMin);            curMin = min(curMin, prices[i]);        }        postProfit[n-1] = 0;        int curMax = prices[n-1];        for (int i = n - 2; i >= 0; --i) {            postProfit[i] = max(postProfit[i+1], curMax - prices[i]);            curMax = max(curMax, prices[i]);        }        int maxProfit = 0;        for (int i = 0; i < n; ++i) {            maxProfit = max(maxProfit, preProfit[i] + postProfit[i]);        }        return maxProfit;    }};