LeetCode（123） Best Time to Buy and Sell Stock III

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).

Subscribe to see which companies asked this question

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