/*** 题目：https://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/* 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).O(n^2)的算法非常easy想到：找寻一个点j，将原来的price[0..n-1]切割为price[0..j]和price[j..n-1]，分别求两段的最大profit。进行优化：对于点j+1，求price[0..j+1]的最大profit时，非常多工作是反复的，在求price[0..j]的最大profit中已经做过了。相似于Best Time to Buy and Sell Stock。能够在O(1)的时间从price[0..j]推出price[0..j+1]的最大profit。

可是怎样从price[j..n-1]推出price[j+1..n-1]？反过来思考，我们能够用O(1)的时间由price[j+1..n-1]推出price[j..n-1]。

终于算法： 数组l[i]记录了price[0..i]的最大profit。 数组r[i]记录了price[i..n]的最大profit。 已知l[i]，求l[i+1]是简单的，相同已知r[i]，求r[i-1]也非常easy。 最后，我们再用O(n)的时间找出最大的l[i]+r[i]，即为题目所求。 * 參考：http://blog.csdn.net/pickless/article/details/12034365 * */ class Solution { public: int maxProfit(vector<int> &prices) { const int size = prices.size(); if(size <= 1){ return 0; } int profitLeft[size]; int profitRight[size]; int profit = 0; int i = 0; int min = 0; int max = 0; memset(profitLeft, 0, sizeof(int) * size); memset(profitRight, 0, sizeof(int) * size); //store the maxProfit into profitLeft[i], from left to right //profit = 0; min = prices[0]; for(i = 1; i < size; i++){ //profitLeft[i] = (prices[i] - min) > profitLeft[i -1] ?

(prices[i] - min) : profitLeft[i - 1]; //min = prices[i] < min ? prices[i] : min; if(prices[i] - min > profitLeft[i -1]){//注意，此处不要写成if(prices[i] - min > profit){ profitLeft[i] = prices[i] - min; } else { profitLeft[i] = profitLeft[i - 1]; } if(prices[i] < min){ min = prices[i]; } } //store the maxProfit into profitRight[i], from right to left // profit = 0; max = prices[size - 1]; for(i = size - 2; i >= 0; i--){ if(max - prices[i] > profitRight[i + 1]){//注意，此处不要写成if(max - prices[i] > profit){ profitRight[i] = max - prices[i]; } else { profitRight[i] = profitRight[i + 1]; } if(prices[i] > max){ max = prices[i]; } } profit = profitRight[0] + profitLeft[0]; for(i = 1; i < size; i++){ if(profitRight[i] + profitLeft[i] > profit){ profit = profitRight[i] + profitLeft[i]; } } return profit; } };