题目

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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分析

紧接着121 和122 的另外一道题目，此次要求只能进行两次买卖交易，求最大利润。

一篇很好的分析文章，参考博客。

第一步扫描，先计算出子序列[0,…,i]中的最大利润，用一个数组保存下来，那么时间是O(n)。

计算方法也是利用第一个问题的计算方法。

第二步是逆向扫描，计算子序列[i,…,n-1]上的最大利润，这一步同时就能结合上一步的结果计算最终的最大利润了，这一步也是O(n)。

第三步，求[0,i]的最大利润与[i,n-1]的最大利润之和的最大值，所以最后算法的复杂度就是O(n)的。

AC代码

class Solution {

public:

    int maxProfit(vector<int>& prices) {

        if (prices.empty())

            return 0;

        int n = prices.size();

        vector<int> profits(n, 0), profits_reverse(n,0);

        //正向遍历，profits[i]表示 prices[0...i]内做一次交易的最大收益.

        int low = prices[0] , cur_profit = 0;

        for (int i = 1; i < n; ++i)

        {

            if (prices[i] < low)

            {

                low = prices[i];

            }

            else{

                if (cur_profit < prices[i] - low)

                    cur_profit = prices[i] - low;

            }

            profits[i] = cur_profit;

        }//for

        //逆向遍历, profits_reverse[i]表示 prices[i...n-1]内做一次交易的最大收益.

        //当前最大价格

        int high = prices[n - 1];

        cur_profit = 0;

        for (int i = n - 2; i >= 0; --i)

        {

            if (prices[i] > high)

                high = prices[i];

            else{

                if (cur_profit < high - prices[i])

                    cur_profit = high - prices[i];

            }//else

            profits_reverse[i] = cur_profit;

        }

        int max_profile = 0;

        for (int i = 0; i < n; i++)

        {

            if ((profits[i] + profits_reverse[i]) > max_profile)

                max_profile = profits[i] + profits_reverse[i];

        }

        return max_profile;

    }

};

GitHub测试程序源码